Exercise 19.5
1. $\mathop \smallint \limits_0^1 \left( {{{\rm{x}}^2} + 5} \right).{\rm{dx}}$
Solution:
Or, $\mathop \smallint \limits_0^1 \left( {{{\rm{x}}^2} + 5} \right).{\rm{dx}}$ = $\left[ {\frac{{{{\rm{x}}^3}}}{3} + 5{\rm{x}}} \right]_0^1$ = $\frac{1}{3}$ + 5 – 0 = $5\frac{1}{3}$.
2. $\mathop \smallint \limits_1^2 \left( {2{{\rm{x}}^2} + 3{\rm{x}} + 4} \right).{\rm{dx}}$
Solution:
Or, $\mathop \smallint \limits_1^2 \left( {2{{\rm{x}}^2} + 3{\rm{x}} + 4} \right).{\rm{dx}}$ = $\left[ {\frac{{2{{\rm{x}}^3}}}{3} + \frac{{3{{\rm{x}}^2}}}{2} + 4{\rm{x}}} \right]_1^2$
= $\left\{ {\frac{2}{3}{\rm{*}}{{\left( 2 \right)}^3} + \frac{3}{2}{{\left( 2 \right)}^2} + 4{\rm{*}}2} \right\}$ – $\left\{ {\frac{2}{3}{\rm{*}}1 + \frac{3}{2}{\rm{*}}1 + 4{\rm{*}}1} \right\}$
= $\frac{{16}}{3} + 6 + 8 - \frac{2}{3} - \frac{2}{3} - 4$ = $13\frac{1}{6}$.
3. $\mathop \smallint \limits_{ - 1}^2 {\left( {{\rm{x}} + 2} \right)^2}.{\rm{dx}}$
Solution:
Or, $\mathop \smallint \limits_{ - 1}^2 {\left( {{\rm{x}} + 2} \right)^2}.{\rm{dx}}$ = $\left[ {\frac{{{{\left( {{\rm{x}} + 2} \right)}^3}}}{3}} \right]_{ - 1}^2$
= $\frac{{{{\left( {2 + 2} \right)}^3}}}{3} - \frac{{{{\left( { - 1 + 2} \right)}^3}}}{3}$ = $\frac{{64}}{3} - \frac{1}{3}$ = 21.
4. $\mathop \smallint \limits_0^{ - 1} \frac{{{\rm{dx}}}}{{{\rm{x}} + 2}}$
Solution:
Or, $\mathop \smallint \limits_0^{ - 1} \frac{{{\rm{dx}}}}{{{\rm{x}} + 2}}$ = $\left[ {{\rm{log}}\left( {{\rm{x}} + 2} \right)} \right]_0^{ - 1}$
= log(–1 + 2) – log(0 + 2) = 0 – log 2 = –log2.
5. $\mathop \smallint \limits_0^1 {{\rm{x}}^3}\sqrt {1 + 2{{\rm{x}}^4}} $.dx
Solution:
Let I = $\mathop \smallint \limits_0^1 {{\rm{x}}^3}\sqrt {1 + 2{{\rm{x}}^4}} $.dx
Put 1 + 2x4 = y ⇒ 8x3.dx = 8y ⇒ x3.dx = $\frac{1}{8}$dy.
When x = 0, y = 1; x = 1, y = 3.
I = $\mathop \smallint \limits_1^3 \sqrt {\rm{y}} \frac{1}{8}$.dy = $\frac{1}{8}\mathop \smallint \limits_1^3 {{\rm{y}}^{\frac{1}{2}}}$.dy = $\left[ {\frac{1}{8}\frac{{{{\rm{y}}^{\frac{3}{2}}}}}{{\frac{3}{2}}}} \right]_1^3$
= $\frac{1}{{12}}$ [(3)3/2 – (1)3/2] = $\frac{1}{{12}}$(3$\sqrt 3 $ – 1).
6. $\mathop \smallint \limits_1^2 {{\rm{e}}^{2{{\rm{x}}^2} - 1}}.2{\rm{x}}$.dx
Solution:
Let I = $\mathop \smallint \limits_1^2 {{\rm{e}}^{2{{\rm{x}}^2} - 1}}.2{\rm{x}}$.dx
Put 2x2 – 1 = y ⇒ 4x.dx = dy ⇒ 2x.dx = $\frac{1}{2}$dy.
When x = 1, y = 1; x = 2, y = 7.
I = $\mathop \smallint \limits_1^7 {{\rm{e}}^{\rm{y}}}.\frac{1}{2}{\rm{dy}}$
= $\frac{1}{2}$.${\rm{\: }}\mathop \smallint \limits_1^7 {{\rm{e}}^{\rm{y}}}{\rm{dy}}$ = $\left[ {\frac{1}{2}{{\rm{e}}^{\rm{y}}}} \right]_1^7$ = $\frac{1}{2}$(e7 – e).
7. $\mathop \smallint \limits_0^{\rm{a}} \frac{{\rm{x}}}{{{{\left( {{{\rm{a}}^2} + {{\rm{x}}^2}} \right)}^{\frac{3}{2}}}}}$.dx
Solution:
Let I = $\mathop \smallint \limits_0^{\rm{a}} \frac{{\rm{x}}}{{{{\left( {{{\rm{a}}^2} + {{\rm{x}}^2}} \right)}^{\frac{3}{2}}}}}$.dx
Put a2 + x2 = y ⇒ 2x.dx = dy ⇒ x.dx = $\frac{1}{2}$.dy
When x = 0, y = a2; x = a, y = 2a2.
I = $\mathop \smallint \limits_{\rm{a}}^{2{{\rm{a}}^2}} \frac{{\frac{1}{2}.{\rm{dy}}}}{{{{\rm{y}}^{\frac{3}{2}}}}}$ = $\frac{1}{2}\mathop \smallint \limits_{\rm{a}}^{2{{\rm{a}}^2}} {{\rm{y}}^{ - \frac{3}{2}}}.{\rm{dy}}$ = $\frac{1}{2}{\left[ {\frac{{{{\rm{y}}^{ - \frac{3}{2} + 1}}}}{{ - \frac{1}{2}}}} \right]^{2{{\rm{a}}^2}}}_{\rm{a}}^2$
= $ - {\left[ {{{\rm{y}}^{ - \frac{1}{2}}}} \right]^{2{{\rm{a}}^2}}}_{\rm{a}}^2$ = $ - {\left[ {\frac{1}{{\sqrt {\rm{y}} }}} \right]^{2{{\rm{a}}^2}}}_{\rm{a}}^2$ = $ - \left[ {\frac{1}{{\sqrt {2{{\rm{a}}^2}} }} - \frac{1}{{\sqrt {{{\rm{a}}^2}} }}} \right]$
= $\frac{1}{{\rm{a}}} - \frac{1}{{\sqrt 2 {\rm{a}}}}$
= $\frac{{\sqrt 2 - 1}}{{\sqrt 2 {\rm{a}}}}$.
8. $\mathop \smallint \limits_0^1 \frac{{2{\rm{x}}.{\rm{dx}}}}{{{{\rm{x}}^2} + 3}}$
Solution:
Let I = $\mathop \smallint \limits_0^1 \frac{{2{\rm{x}}.{\rm{dx}}}}{{{{\rm{x}}^2} + 3}}$
Put x2 + 3 = y ⇒ 2x.dx = dy
When x = 0, y = 3; x = 1, y = 4.
I = $\mathop \smallint \limits_3^4 \frac{{{\rm{dy}}}}{{\rm{y}}}$ = $\left[ {{\rm{log}}} \right]_3^4$ = log4 – log3.
9. $\mathop \smallint \limits_0^{\rm{a}} {\rm{x}}\sqrt {{{\rm{a}}^2} + {{\rm{x}}^2}.{\rm{dx}}} $
Solution:
Let I = $\mathop \smallint \limits_0^{\rm{a}} {\rm{x}}\sqrt {{{\rm{a}}^2} + {{\rm{x}}^2}.{\rm{dx}}} $
Put a2 + x2 = y ⇒ 2x.dx = dy ⇒ x.dx = $\frac{1}{2}$.dy
When x = 0, y = a2; x = a, y = 2a2.
I = $\mathop \smallint \limits_{{{\rm{a}}^2}}^{2{{\rm{a}}^2}} \sqrt {\rm{y}} .\frac{1}{2}.{\rm{dy}}$ = $\frac{1}{2}\mathop \smallint \limits_{{{\rm{a}}^2}}^{2{{\rm{a}}^2}} {{\rm{y}}^{\frac{1}{2}}}$.dy
= $\frac{1}{2}$$\left[ {\frac{{{{\rm{y}}^{\frac{3}{2}}}}}{{\frac{3}{2}}}} \right]_{{{\rm{a}}^2}}^{2{{\rm{a}}^2}}$ = $\frac{1}{2}\left[ {{{\left( {2{{\rm{a}}^2}} \right)}^{\frac{3}{2}}} - {{\left( {{{\rm{a}}^2}} \right)}^{\frac{3}{2}}}} \right]$
= $\frac{1}{3}\left( {2\sqrt 2 {{\rm{a}}^3} - {{\rm{a}}^3}} \right)$ = $\frac{1}{3}\left( {2\sqrt 2 - 1} \right){{\rm{a}}^3}$.
10. $\mathop \smallint \limits_{ - \frac{{\rm{\pi }}}{3}}^{\frac{{\rm{\pi }}}{3}} {\rm{cost}}.{\rm{dt}}$
Solution:
Or, $\mathop \smallint \limits_{ - \frac{{\rm{\pi }}}{3}}^{\frac{{\rm{\pi }}}{3}} {\rm{cost}}.{\rm{dt}}$
= $\left[ {{\rm{sint}}} \right]_{\frac{{\rm{\pi }}}{3}}^{\frac{{\rm{\pi }}}{3}}$ = sin $\frac{{\rm{\pi }}}{3}$ – $\sin \left( { - \frac{{\rm{\pi }}}{3}} \right)$ = $\frac{{\sqrt 3 }}{2} + \frac{{\sqrt 3 }}{2}$ = $\sqrt 3 $.
11. $\mathop \smallint \limits_0^{\frac{{\rm{\pi }}}{4}} {\sec ^2}\theta .{\rm{d}}\theta $
Solution:
Or, $\mathop \smallint \limits_0^{\frac{{\rm{\pi }}}{4}} {\sec ^2}\theta .{\rm{d}}\theta $
= $\left[ {{\rm{tan}}\theta } \right]_0^{\frac{{\rm{\pi }}}{4}}$ = tan $\frac{{\rm{\pi }}}{4}$ – tan0 = 1 – 0 = 1.
12. $\mathop \smallint \limits_0^2 {\cos ^2}{\rm{\pi x}}.{\rm{dx}}$
Solution:
Or, $\mathop \smallint \limits_0^2 {\cos ^2}{\rm{\pi x}}.{\rm{dx}}$
= $\frac{1}{2}$$\mathop \smallint \limits_0^1 2.{\cos ^2}{\rm{\pi x}}$ = $\frac{1}{2}\mathop \smallint \limits_0^1 \left( {1 + {\rm{cos}}2{\rm{\pi x}}} \right)$.dx
= $\frac{1}{2}$$\left[ {{\rm{x}} + \frac{{{\rm{sin}}2{\rm{\pi x}}}}{{2{\rm{\pi }}}}} \right]_0^1$ = $\frac{1}{2}\left\{ {\left( {1 + \frac{{{\rm{sin}}2{\rm{\pi }}}}{{\rm{\pi }}}} \right) - 0} \right\}$ = $\frac{1}{2}$(1 + 0) = $\frac{1}{2}$.
13. $\mathop \smallint \limits_0^{\frac{{\rm{\pi }}}{4}} {\sin ^2}{\rm{x}}.{\rm{dx}}$
Solution:
Or, $\mathop \smallint \limits_0^{\frac{{\rm{\pi }}}{4}} {\sin ^2}{\rm{x}}.{\rm{dx}}$
= $\frac{1}{2}$$\mathop \smallint \limits_0^{\frac{{\rm{\pi }}}{4}} 2.{\sin ^2}{\rm{x}}.{\rm{dx}}$ = $\frac{1}{2}\mathop \smallint \limits_0^{\frac{{\rm{\pi }}}{4}} \left( {1 - {\rm{cos}}2{\rm{x}}} \right)$.dx
= $\frac{1}{2}$$\left[ {{\rm{x}} - \frac{{{\rm{sin}}2{\rm{x}}}}{2}} \right]_0^{\frac{{\rm{\pi }}}{4}}$ = $\frac{1}{2}\left( {\frac{{\rm{\pi }}}{4} - \frac{{\sin \frac{{\rm{\pi }}}{2}}}{2} - 0} \right)$
= $\frac{1}{2}$($\frac{{\rm{\pi }}}{4}$ – $\frac{1}{2}$) = $\frac{{\rm{\pi }}}{8} - \frac{1}{4}$.
14. $\mathop \smallint \limits_0^{\frac{{\rm{\pi }}}{6}} {\cos ^3}{\rm{x}}.{\rm{dx}}$
Solution:
Or, $\mathop \smallint \limits_0^{\frac{{\rm{\pi }}}{6}} {\cos ^3}{\rm{x}}.{\rm{dx}}$
= $\frac{1}{4}$$\mathop \smallint \limits_0^{\frac{{\rm{\pi }}}{6}} 4.{\cos ^3}{\rm{x}}.{\rm{dx}}$ = $\frac{1}{4}\mathop \smallint \limits_0^{\frac{{\rm{\pi }}}{6}} \left( {{\rm{cos}}3{\rm{x}} + 3{\rm{cosx}}} \right)$.dx (cos3x = 4cos3x – 3cosx)
= $\frac{1}{4}$$\left[ {\frac{{{\rm{sin}}3{\rm{x}}}}{3} + 3{\rm{sinx}}} \right]_0^{\frac{{\rm{\pi }}}{6}}$ = $\frac{1}{4}\left( {\frac{{\sin \frac{{\rm{\pi }}}{2}}}{3} + 3\sin \frac{{\rm{\pi }}}{6} - 0} \right)$
= $\frac{1}{4}\left[ {\frac{1}{3} + \frac{{3.1}}{2}} \right]$ = $\frac{{11}}{{24}}$.
15. $\mathop \smallint \limits_0^{\frac{{\rm{\pi }}}{2}} {\sin ^3}{\rm{x}}.{\rm{dx}}$
Solution:
Or, $\mathop \smallint \limits_0^{\frac{{\rm{\pi }}}{2}} {\sin ^3}{\rm{x}}.{\rm{dx}}$
= $\frac{1}{4}$$\mathop \smallint \limits_0^{\frac{{\rm{\pi }}}{4}} 4.{\sin ^3}{\rm{x}}.{\rm{dx}}$ = $\frac{1}{4}\mathop \smallint \limits_0^{\frac{{\rm{\pi }}}{2}} \left( {3{\rm{sinx}} - {\rm{sin}}3{\rm{x}}} \right)$.dx (sin3x = 3sinx – 4sin3x)
= $\frac{1}{4}$$\left[ { - 3{\rm{cosx}} - \frac{{ - {\rm{cos}}3{\rm{x}}}}{3}} \right]_0^{\frac{{\rm{\pi }}}{2}}$ = $\frac{1}{4}\left[ { - 3{\rm{cosx}} + \frac{{{\rm{cos}}3{\rm{x}}}}{3}} \right]_0^{\frac{{\rm{\pi }}}{2}}$
= $\frac{1}{4}$$\left[ {\left\{ { - 3\cos \frac{{\rm{\pi }}}{2} + \frac{{\cos \frac{{3{\rm{\pi }}}}{2}}}{3}} \right\} - \left\{ { - 3{\rm{cos}}0 + \frac{{{\rm{cos}}0}}{3}} \right\}} \right]$
= $\frac{1}{4}\left[ {0 + 3.1 - \frac{1}{3}} \right]$ = $\frac{2}{3}$.
16. $\mathop \smallint \limits_0^{\frac{{\rm{\pi }}}{3}} \frac{{{\rm{dx}}}}{{2{{\cos }^2}\frac{{\rm{x}}}{2}}}$
Solution:
Or, $\mathop \smallint \limits_0^{\frac{{\rm{\pi }}}{3}} \frac{{{\rm{dx}}}}{{2{{\cos }^2}\frac{{\rm{x}}}{2}}}$ = $\frac{1}{2}\mathop \smallint \limits_0^{\frac{{\rm{\pi }}}{3}} {\sec ^2}\frac{{\rm{x}}}{2}.{\rm{dx}}$
= $\frac{1}{2}\left[ {\frac{{\tan \frac{{\rm{x}}}{2}}}{{\frac{1}{2}}}} \right]_0^{\frac{{\rm{\pi }}}{3}}$ = $\left[ {\tan \frac{{\rm{x}}}{2}} \right]_0^{\frac{{\rm{\pi }}}{3}}$ = tan $\frac{{\rm{\pi }}}{6}$ – 0 = $\frac{1}{{\sqrt 3 }}$.
17. $\mathop \smallint \limits_{\frac{{\rm{\pi }}}{2}}^{\frac{{\rm{\pi }}}{6}} \frac{{{\rm{dx}}}}{{\left( {1 - {\rm{cos}}2{\rm{x}}} \right)}}$
Solution:
Or, $\mathop \smallint \limits_{\frac{{\rm{\pi }}}{2}}^{\frac{{\rm{\pi }}}{6}} \frac{{{\rm{dx}}}}{{\left( {1 - {\rm{cos}}2{\rm{x}}} \right)}}$ = $\mathop \smallint \limits_{\frac{{\rm{\pi }}}{2}}^{\frac{{\rm{\pi }}}{6}} \frac{{{\rm{dx}}}}{{2{{\sin }^2}{\rm{x}}}}$ = $\frac{1}{2}\mathop \smallint \limits_{\frac{{\rm{\pi }}}{2}}^{\frac{{\rm{\pi }}}{6}} {\rm{cose}}{{\rm{c}}^2}{\rm{x}}$.dx
= $\frac{1}{2}\left[ {\left( { - {\rm{cotx}}} \right)} \right]_{\frac{{\rm{\pi }}}{2}}^{\frac{{\rm{\pi }}}{6}}$ = $ - \frac{1}{2}\left[ {\cot \frac{{\rm{\pi }}}{6} - \cot \frac{{\rm{\pi }}}{2}} \right]$ = $ - \frac{1}{2}\left[ {\sqrt 3 - 0} \right]$ = $ - \frac{{\sqrt 3 }}{2}$.
18. $\mathop \smallint \limits_0^{\frac{{\rm{\pi }}}{4}} \frac{{{\rm{dx}}}}{{\left( {1 - {\rm{sinx}}} \right)}}$
Solution:
Or, $\mathop \smallint \limits_0^{\frac{{\rm{\pi }}}{4}} \frac{{{\rm{dx}}}}{{\left( {1 - {\rm{sinx}}} \right)}}$
= $\mathop \smallint \limits_0^{\frac{{\rm{\pi }}}{4}} \frac{{{\rm{dx}}}}{{\left( {1 - {\rm{sinx}}} \right)}}{\rm{*}}\frac{{1 + {\rm{sinx}}}}{{1 + {\rm{sinx}}}}$
= $\mathop \smallint \limits_0^{\frac{{\rm{\pi }}}{4}} \frac{{1 + {\rm{sinx}}}}{{1 - {{\sin }^2}{\rm{x}}}}$
= $\mathop \smallint \limits_0^{\frac{{\rm{\pi }}}{4}} \frac{{1 + {\rm{sinx}}}}{{{{\cos }^2}{\rm{x}}}}$
= $\mathop \smallint \limits_0^{\frac{{\rm{\pi }}}{4}} [\frac{1}{{{{\cos }^2}{\rm{x}}}} + \frac{{{\rm{sinx}}}}{{{{\cos }^2}{\rm{x}}}}]$.dx
= ${\rm{\: \: \: \: }}\mathop \smallint \limits_0^{\frac{{\rm{\pi }}}{4}} [{\sec ^2}{\rm{x}} + {\rm{tanx}}.{\rm{secx}}].{\rm{dx}}$ = $\left[ {{\rm{tanx}} + {\rm{secx}}} \right]_0^{\frac{{\rm{\pi }}}{4}}$
= $\left( {\tan \frac{{\rm{\pi }}}{4} + \sec \frac{{\rm{\pi }}}{4}} \right)$ – (tan0 + sec0) = 1 + $\sqrt 2 $ – 0 – 1 = $\sqrt 2 $.
19. $\mathop \smallint \limits_0^{\frac{{\rm{\pi }}}{2}} \frac{{{\rm{dx}}}}{{1 + {\rm{sinx}}}}$
Solution:
I = $\mathop \smallint \limits_0^{\frac{{\rm{\pi }}}{2}} \frac{{{\rm{dx}}}}{{1 + {\rm{sinx}}}}$
Put x = $\frac{{\rm{\pi }}}{2}$ – y ⇒ dx = –dy.
When x = 0, y = $\frac{{\rm{\pi }}}{2}$; x = $\frac{{\rm{\pi }}}{2}$, y = 0.
Or, I = $\mathop \smallint \limits_0^{\frac{{\rm{\pi }}}{2}} \frac{{ - {\rm{dy}}}}{{1 + {\rm{sin}}\left( {\frac{{\rm{\pi }}}{2} - {\rm{y}}} \right)}}$
= $\mathop \smallint \limits_0^{\frac{{\rm{\pi }}}{2}} \frac{{ - {\rm{dy}}}}{{1 + {\rm{cosy}}}}$
= $\mathop \smallint \limits_0^{\frac{{\rm{\pi }}}{2}} \frac{{ - {\rm{dy}}}}{{2{{\cos }^2}\frac{{\rm{y}}}{2}}}$
= $\frac{1}{2}\mathop \smallint \limits_0^{\frac{{\rm{\pi }}}{2}} {\sec ^2}\frac{{\rm{y}}}{2}$.dy
= $ - \frac{1}{2}\left[ {\frac{{\tan \frac{{\rm{y}}}{2}}}{{\frac{1}{2}}}} \right]_{\frac{{\rm{\pi }}}{2}}^0$ = $ - \left[ {{\rm{tan}}0 - \tan \frac{{\rm{\pi }}}{4}} \right]$ = $ - \left( {0 - 1} \right)$ = 1.
20. $\mathop \smallint \limits_0^{\rm{\pi }} \sqrt {1 + {\rm{cosx}}} $.dx
Solution:
Or, I = $\mathop \smallint \limits_0^{\rm{\pi }} \sqrt {1 + {\rm{cosx}}} $.dx
= $\mathop \smallint \limits_0^{\rm{\pi }} \sqrt {2{{\cos }^2}\frac{{\rm{x}}}{2}} $.dx
= $\sqrt 2 $. $\mathop \smallint \limits_0^{\rm{\pi }} \cos \frac{{\rm{x}}}{2}$.dx = $\sqrt 2 \left[ {\frac{{\sin \frac{{\rm{x}}}{2}}}{{\frac{1}{2}}}} \right]_0^{\rm{\pi }}$ = $2\sqrt 2 \left[ {\sin \frac{{\rm{x}}}{2}} \right]_0^{\rm{\pi }}$
= $2\sqrt 2 $$\left[ {\sin \frac{{\rm{\pi }}}{2} - {\rm{sin}}0} \right]$ = $2\sqrt 2 $[1 – 0] = $2\sqrt 2 $.
21. $\mathop \smallint \limits_0^{\frac{{\rm{\pi }}}{2}} \sqrt {1 + {\rm{sinx}}} $.dx
Solution:
Or, I = $\mathop \smallint \limits_0^{\frac{{\rm{\pi }}}{2}} \sqrt {1 + {\rm{sinx}}} $.dx
= $\mathop \smallint \limits_0^{\frac{{\rm{\pi }}}{2}} \sqrt {({{\sin }^2}\left( {\frac{{\rm{x}}}{2}} \right) + {{\cos }^2}\left( {\frac{{\rm{x}}}{2}} \right) + 2.{\rm{sin}}\frac{{\rm{x}}}{2}.{\rm{cos}}\frac{{\rm{x}}}{2}} $.dx
= $\mathop \smallint \limits_0^{\frac{{\rm{\pi }}}{2}} \sqrt {{{\left( {{\rm{sin}}\frac{{\rm{x}}}{2} + {\rm{cos}}\frac{{\rm{x}}}{2}} \right)}^2}} $.dx
= $\mathop \smallint \limits_0^{\frac{{\rm{\pi }}}{2}} \left( {{\rm{sin}}\frac{{\rm{x}}}{2} + {\rm{cos}}\frac{{\rm{x}}}{2}} \right)$.dx
= $\left[ { - \frac{{\cos \frac{{\rm{x}}}{2}}}{{\frac{1}{2}}} + \frac{{{\rm{sin}}\frac{{\rm{x}}}{2}}}{{\frac{1}{2}}}} \right]_0^{\frac{{\rm{\pi }}}{2}}$ = 2$\left( { - {\rm{cos}}\frac{{\rm{x}}}{2} + {\rm{sin}}\frac{{\rm{x}}}{2}} \right)_0^{\frac{{\rm{\pi }}}{2}}$
= 2$\left\{ {\left( { - {\rm{cos}}\frac{{\rm{\pi }}}{4} + {\rm{sin}}\frac{{\rm{\pi }}}{4}} \right) - \left( { - {\rm{cos}}0 + {\rm{sin}}0} \right)} \right\}{\rm{\: }}$
= $2\left\{ { - \frac{1}{{\sqrt 2 }} + \frac{1}{{\sqrt 2 }} + 1 - 0} \right\}$ = 2.
22. $\mathop \smallint \limits_0^{\frac{{\rm{\pi }}}{3}} {\rm{sin}}2\theta .{\rm{cos}}\theta $.dθ
Solution:
Or, $\mathop \smallint \limits_0^{\frac{{\rm{\pi }}}{3}} {\rm{sin}}2\theta .{\rm{cos}}\theta $.dθ
= $\frac{1}{2}$.2sin2θ.cosθ.dθ
= $\frac{1}{2}$$\mathop \smallint \limits_0^{\frac{{\rm{\pi }}}{3}} \left\{ {\sin \left( {2\theta + \theta } \right) + \sin \left( {2\theta - \theta } \right)} \right\}$.dθ
= $\frac{1}{2}$$\mathop \smallint \limits_0^{\frac{{\rm{\pi }}}{3}} \left( {{\rm{sin}}3\theta + {\rm{sin}}\theta } \right)$.dθ = $\frac{1}{2}\left[ { - \frac{{{\rm{cos}}3\theta }}{3} - {\rm{cos}}\theta } \right]_0^{\frac{{\rm{\pi }}}{3}}$
= $\frac{1}{2}\left[ {\left\{ { - \frac{{{\rm{cos\pi }}}}{3} - \cos \frac{{\rm{\pi }}}{3}} \right\} - \left\{ { - \frac{{{\rm{cos}}0}}{3} - {\rm{coso}}} \right\}} \right]$
= $\frac{1}{2}\left[ {\frac{1}{3} - \frac{1}{2} + \frac{1}{3} + 1} \right]$ = $\frac{7}{{12}}$.
23. $\mathop \smallint \limits_0^{\frac{{\rm{\pi }}}{2}} {\rm{cos}}3{\rm{x}}.{\rm{cos}}2{\rm{x}}$.dx
Solution:
Or, $\mathop \smallint \limits_0^{\frac{{\rm{\pi }}}{2}} {\rm{cos}}3{\rm{x}}.{\rm{cos}}2{\rm{x}}$.dx
= $\frac{1}{2}$.${\rm{\: }}\mathop \smallint \limits_0^{\frac{{\rm{\pi }}}{2}} 2.{\rm{cos}}3{\rm{x}}.{\rm{cos}}2{\rm{x}}$.dx
= $\frac{1}{2}$$\mathop \smallint \limits_0^{\frac{{\rm{\pi }}}{2}} \left\{ {\cos \left( {3{\rm{x}} + 2{\rm{x}}} \right) + \cos \left( {3{\rm{x}} - 2{\rm{x}}} \right)} \right\}$.dx
= $\frac{1}{2}$$\mathop \smallint \limits_0^{\frac{{\rm{\pi }}}{2}} \left( {{\rm{cos}}5{\rm{x}} + {\rm{cosx}}} \right)$.dx
= $\frac{1}{2}\left[ {\frac{{{\rm{sin}}5{\rm{x}}}}{5} + {\rm{sinx}}} \right]_0^{\frac{{\rm{\pi }}}{2}}$
= $\frac{1}{2}\left[ {\frac{{\sin \frac{{5{\rm{\pi }}}}{2}}}{5} + \sin \frac{{\rm{\pi }}}{2} - 0} \right]$
= $\frac{1}{2}\left( {\frac{1}{5} + 1} \right)$ = $\frac{3}{5}$.
24. $\mathop \smallint \limits_0^{\frac{{\rm{\pi }}}{4}} {\cos ^3}{\rm{x}}.{\sin ^2}{\rm{x}}$.dx
Solution:
Let I = $\mathop \smallint \limits_0^{\frac{{\rm{\pi }}}{4}} {\cos ^3}{\rm{x}}.{\sin ^2}{\rm{x}}$.dx
= $\mathop \smallint \limits_0^{\frac{{\rm{\pi }}}{4}} {\cos ^2}{\rm{x}}.{\rm{cosx}}.{\sin ^2}{\rm{x}}$.dx
= $\mathop \smallint \limits_0^{\frac{{\rm{\pi }}}{4}} (1 - {\sin ^2}{\rm{x}}).{\sin ^2}{\rm{x}}.{\rm{cosx}}$.dx
Put sinx = y ⇒ cos.dx = dy
When, x = 0, y = 0; x = $\frac{{\rm{\pi }}}{4}$,y = $\frac{1}{{\sqrt 2 }}$.
Or, I = $\mathop \smallint \limits_0^{\frac{1}{{\sqrt 2 }}} \left( {1 - {{\rm{y}}^2}} \right).{{\rm{y}}^2}$.dy = $\mathop \smallint \limits_0^{\frac{1}{{\sqrt 2 }}} \left( {{{\rm{y}}^2} - {{\rm{y}}^4}} \right).{\rm{dy}}$
= $\left[ {\frac{{{{\rm{y}}^3}}}{3} - \frac{{{{\rm{y}}^5}}}{4}} \right]_0^{\frac{1}{{\sqrt 2 }}}$ = $\frac{{{{\left( {\frac{1}{{\sqrt 2 }}} \right)}^3}}}{3} - \frac{{{{\left( {\frac{1}{{\sqrt 2 }}} \right)}^5}}}{5}$
= $\frac{1}{{6\sqrt 2 }} - \frac{1}{{20\sqrt 2 }}$ = $\frac{{10 - 3}}{{60\sqrt 2 }}$ = $\frac{7}{{60\sqrt 2 }}$.
25. $\mathop \smallint \limits_0^{\frac{{\rm{\pi }}}{2}} {\left( {1 + {\rm{cosx}}} \right)^2}$.sinx.dx
Solution:
Let I = $\mathop \smallint \limits_0^{\frac{{\rm{\pi }}}{2}} {\left( {1 + {\rm{cosx}}} \right)^2}$.sinx.dx
Put 1 + cosx = y ⇒ –sinx.dx = dy⇒ sinx.dx = –dy.
When, x = 0, y = 2; x = $\frac{{\rm{\pi }}}{2}$,y = 1.
Or, I = $ - \mathop \smallint \limits_2^1 {{\rm{y}}^2}$.dy
= $\left[ {\frac{{{{\rm{y}}^3}}}{3}} \right]_2^1$ = $ - \left[ {\frac{1}{3} - \frac{8}{3}} \right]$ = $\frac{7}{3}$.
26. $\mathop \smallint \limits_0^{\frac{{\rm{\pi }}}{4}} {\rm{tanx}}$.dx
Solution:
Let I = $\mathop \smallint \limits_0^{\frac{{\rm{\pi }}}{4}} {\rm{tanx}}$.dx
= $\mathop \smallint \limits_0^{\frac{{\rm{\pi }}}{4}} \frac{{{\rm{sinx}}}}{{{\rm{cosx}}}}$.dx
Put cosx = y ⇒ –sinx.dx = dy ⇒ sinx.dx = –dy.
When, x = 0, y = 1; x = $\frac{{\rm{\pi }}}{4}$,y = $\frac{1}{{\sqrt 2 }}$.
Or, I = $\mathop \smallint \limits_1^{\frac{1}{{\sqrt 2 }}} - \frac{{{\rm{dy}}}}{{\rm{y}}}$ = $ - \left[ {{\rm{log}}} \right]_1^{\frac{1}{{\sqrt 2 }}}$ = $ - \left( {\log \frac{1}{{\sqrt 2 }} - {\rm{log}}1} \right)$.
= $ - \left\{ {{\rm{log}}1 - {\rm{log}}\sqrt 2 - {\rm{log}}1} \right\}$ = log$\sqrt 2 $ = $\frac{1}{2}$log2.
27. $\mathop \smallint \limits_0^{\frac{{\rm{\pi }}}{4}} {\tan ^3}{\rm{x}}$.dx
Solution:
Let I = $\mathop \smallint \limits_0^{\frac{{\rm{\pi }}}{4}} {\tan ^3}{\rm{x}}$.dx
= $\mathop \smallint \limits_0^{\frac{{\rm{\pi }}}{4}} {\rm{tanx}}.{\tan ^2}{\rm{x}}$.dx = $\mathop \smallint \limits_0^{\frac{{\rm{\pi }}}{4}} {\rm{tanx}}.\left( {{{\sec }^2}{\rm{x}} - 1} \right)$.dx
= $\mathop \smallint \limits_0^{\frac{{\rm{\pi }}}{4}} {\rm{tanx}}.{\sec ^2}{\rm{x}}$.dx $ - \mathop \smallint \limits_0^{\frac{{\rm{\pi }}}{4}} {\rm{tanx}}$ = I1 – I2.
I1 = $\mathop \smallint \limits_0^{\frac{{\rm{\pi }}}{4}} {\rm{tanx}}.{\sec ^2}{\rm{x}}$.dx ${\rm{\: }}$
Put tanx = y ⇒ sec2x.dx = dy
When, x = 0, y = 0; x = $\frac{{\rm{\pi }}}{4}$,y = 1.
Or, I1 = $\mathop \smallint \limits_0^1 {\rm{y}}.$dy = $\frac{{{{\rm{y}}^2}}}{2}$ = $\frac{1}{2}$.
I2 = $\mathop \smallint \limits_0^{\frac{{\rm{\pi }}}{4}} {\rm{tanx}}.$dx = $\mathop \smallint \limits_0^{\frac{{\rm{\pi }}}{4}} \frac{{{\rm{tanx}}.{\rm{secs}}}}{{{\rm{secx}}}}.$dx
Put secx = z ⇒ secx.tanx.dx = dz.
When x = 0,z = 1; x = $\frac{{\rm{\pi }}}{4}$, z = $\sqrt 2 $
I2 = $\mathop \smallint \limits_0^{\sqrt 2 } \frac{{{\rm{dz}}}}{{\rm{z}}}$ = $\left[ {{\rm{logz}}} \right]_0^{\sqrt 2 }$ = log$\sqrt 2 $ – log1 = $\frac{1}{2}$log2 – 0 = $\frac{1}{2}$log2
I = I1 – I2 = $\frac{1}{2} - \frac{1}{2}$log2 = $\frac{1}{2}$(1 – log2).
28. $\mathop \smallint \limits_{\rm{\pi }}^{\frac{{\rm{\pi }}}{2}} {\rm{x}}.{\rm{cosx}}$.dx
Solution:
Let I = $\mathop \smallint \limits_{\rm{\pi }}^{\frac{{\rm{\pi }}}{2}} {\rm{x}}.{\rm{cosx}}$.dx
= $\mathop \smallint \limits_0^{\frac{{\rm{\pi }}}{4}} {\tan ^2}{\rm{x}}\left( {1 + {{\tan }^2}{\rm{x}}} \right){\sec ^2}{\rm{x}}$.dx
Put tanx = y ⇒ sec2x.dx = dy
When x = 0, y = 0; x = $\frac{{\rm{\pi }}}{4}$, y = 1.
I = $\mathop \smallint \limits_0^1 {{\rm{y}}^2}\left( {1 + {{\rm{y}}^2}} \right)$.dy = $\mathop \smallint \limits_0^1 \left( {{{\rm{y}}^2} + {{\rm{y}}^4}} \right)$.dy
= $\left[ {\frac{{{{\rm{y}}^3}}}{3} + \frac{{{{\rm{y}}^5}}}{5}} \right]_0^1$ = $\frac{1}{3}$ + $\frac{1}{5}$ = $\frac{8}{{15}}$.
29. $\mathop \smallint \limits_1^{\rm{e}} {\rm{x}}.{\rm{logx}}$.dx
Solution:
Or, $\mathop \smallint \limits_1^{\rm{e}} {\rm{x}}.{\rm{logx}}$.dx
= logx$\mathop \smallint \nolimits {\rm{x}}$.dx – $\mathop \smallint \nolimits \{ \frac{{\rm{d}}}{{{\rm{dx}}}}\left( {{\rm{logx}}} \right)\mathop \smallint \nolimits {\rm{x}}.{\rm{dx}}\} $.dx
= logx.$\frac{{{{\rm{x}}^2}}}{2}$ – $\mathop \smallint \nolimits \frac{1}{{\rm{x}}}.\frac{{{{\rm{x}}^2}}}{2}$.dx = $\frac{{{{\rm{x}}^2}}}{2}$logx – $\frac{1}{2}\mathop \smallint \nolimits {\rm{x}}.$dx
= $\frac{{{{\rm{x}}^2}}}{2}$.logx – $\frac{1}{2}.\frac{{{{\rm{x}}^2}}}{2}$ = $\frac{{{{\rm{x}}^2}}}{2}$logx – $\frac{{{{\rm{x}}^2}}}{4}$.
$\mathop \smallint \limits_1^{\rm{e}} {\rm{x}}.{\rm{logx}}$.dx = $\left[ {\frac{{{{\rm{x}}^2}}}{2}{\rm{logx}} - \frac{{{{\rm{x}}^2}}}{4}} \right]_1^{\rm{e}}$
= $\left( {\frac{{{{\rm{e}}^2}}}{2}.{\rm{loge}} - \frac{{{{\rm{e}}^2}}}{4}} \right) - \left( {\frac{1}{2}{\rm{log}}1 - \frac{1}{4}} \right)$
= $\frac{{{{\rm{e}}^2}}}{2} - \frac{{{{\rm{e}}^2}}}{4} - 0 + \frac{1}{4}$ = $\frac{{2{{\rm{e}}^2} - {{\rm{e}}^2} + 1}}{4}$ = $\frac{1}{4}$(e2 + 1).
30. $\mathop \smallint \limits_0^1 {\rm{x}}.{{\rm{e}}^{\rm{x}}}$.dx
Solution:
Or, $\mathop \smallint \limits_0^1 {\rm{x}}.{{\rm{e}}^{\rm{x}}}$.dx
= x$\mathop \smallint \nolimits {{\rm{e}}^{\rm{x}}}$.dx – $\mathop \smallint \nolimits \{ \frac{{\rm{d}}}{{{\rm{dx}}}}\left( {\rm{x}} \right)\mathop \smallint \nolimits {{\rm{e}}^{\rm{x}}}.{\rm{dx}}\} $.dx
= x.ex – $\mathop \smallint \nolimits 1.{{\rm{e}}^{\rm{x}}}$.dx = x.ex – ex.
Or, $\mathop \smallint \limits_0^1 {\rm{x}}.{{\rm{e}}^{\rm{x}}}$.dx = $\left[ {{\rm{x}}{{\rm{e}}^{\rm{x}}} - {{\rm{e}}^{\rm{x}}}} \right]_0^1$ = (1.e1 – e1) – (0 – e0) = 1.
31. $\mathop \smallint \limits_1^{\rm{e}} {{\rm{x}}^2}.{\rm{logx}}$.dx
Solution:
Or, $\mathop \smallint \limits_1^{\rm{e}} {{\rm{x}}^2}.{\rm{logx}}$.dx
= logx$\mathop \smallint \nolimits {{\rm{x}}^2}$.dx – $\mathop \smallint \nolimits \{ \frac{{\rm{d}}}{{{\rm{dx}}}}\left( {{\rm{logx}}} \right)\mathop \smallint \nolimits {{\rm{x}}^2}.{\rm{dx}}\} $.dx
= logx$\frac{{{{\rm{x}}^3}}}{3}$ – $\mathop \smallint \nolimits \frac{1}{{\rm{x}}}.\frac{{{{\rm{x}}^3}}}{3}$.dx = $\frac{{{{\rm{x}}^3}}}{3}$logx – $\frac{1}{3}\mathop \smallint \nolimits {{\rm{x}}^2}$.dx
= $\frac{{{{\rm{x}}^3}}}{3}$logx – $\frac{1}{3}$.$\frac{{{{\rm{x}}^3}}}{3}$ = $\frac{1}{3}$x3.logx – $\frac{1}{9}$x3
Or, $\mathop \smallint \limits_1^{\rm{e}} {{\rm{x}}^2}$.logx.dx = $\left[ {\frac{1}{3}{{\rm{x}}^3}{\rm{logx}} - \frac{1}{9}{{\rm{x}}^3}} \right]_1^{\rm{e}}$
= $\left( {\frac{1}{3}{{\rm{e}}^3}{\rm{loge}} - \frac{1}{9}{{\rm{e}}^3}} \right) - \left( {0 - \frac{1}{9}} \right)$
= $\frac{1}{3}$e3 – $\frac{1}{9}$e3 + $\frac{1}{9}$ = $\frac{{3{{\rm{e}}^3} - {{\rm{e}}^3} + 1}}{9}$ = $\frac{{2{{\rm{e}}^3} + 1}}{9}$.
32. $\mathop \smallint \limits_{\rm{\pi }}^{\frac{{\rm{\pi }}}{2}} .{\rm{x}}.{\rm{cosx}}$.dx
Solution:
Or, $\mathop \smallint \limits_{\rm{\pi }}^{\frac{{\rm{\pi }}}{2}} .{\rm{x}}.{\rm{cosx}}$.dx
= x$\mathop \smallint \nolimits {\rm{cosx}}$.dx – $\mathop \smallint \nolimits \{ \frac{{\rm{d}}}{{{\rm{dx}}}}\left( {\rm{x}} \right)\mathop \smallint \nolimits {\rm{cosx}}.{\rm{dx}}\} $.dx
= x.sinx –$\mathop \smallint \nolimits \{ \frac{{\rm{d}}}{{{\rm{dx}}}}\left( {\rm{x}} \right)\mathop \smallint \nolimits {\rm{cosx}}.{\rm{dx}}\} $.dx
= xsinx – $\mathop \smallint \nolimits 1.{\rm{sinx}}$.dx = x.sinx – (– cosx) = xsinx + cosx.
Or, $\mathop \smallint \limits_{\rm{\pi }}^{\frac{{\rm{\pi }}}{2}} {\rm{x}}.{\rm{cosx}}$.dx = $\left[ {{\rm{xsinx}} + {\rm{cosx}}} \right]_{\rm{\pi }}^{\frac{{\rm{\pi }}}{2}}$
= $\left( {\frac{{\rm{\pi }}}{2}.\sin \frac{{\rm{\pi }}}{2} + \cos \frac{{\rm{\pi }}}{2}} \right) - \left( {{\rm{\pi }}.{\rm{sin\pi }} + {\rm{cos\pi }}} \right)$
= $\frac{{\rm{\pi }}}{2}$.1 + 0 – 0 – (–1) = $\frac{{\rm{\pi }}}{2}$ + 1.
33. $\mathop \smallint \limits_0^{\rm{\pi }} .{\rm{x}}.{\sin ^2}{\rm{x}}$.dx
Solution:
Or, $\mathop \smallint \limits_0^{\rm{\pi }} .{\rm{x}}.{\sin ^2}{\rm{x}}$.dx
= $\frac{1}{2}\mathop \smallint \limits_0^{\rm{\pi }} {\rm{x}}(2{\sin ^2}{\rm{x}}).{\rm{dx}}$ = $\frac{1}{2}$$\mathop \smallint \limits_0^{\rm{\pi }} {\rm{x}}$.dx – $\frac{1}{2}$$\mathop \smallint \limits_0^{\rm{\pi }} {\rm{x}}.\left( {1 - {\rm{cos}}2{\rm{x}}} \right)$.dx
= $\frac{1}{2}\mathop \smallint \limits_0^{\rm{\pi }} \left( {{\rm{x}} - {\rm{xcos}}2{\rm{x}}} \right)$.dx = $\frac{1}{2}\mathop \smallint \limits_0^{\rm{\pi }} {\rm{x}}.$dx – $\frac{1}{2}$$\mathop \smallint \limits_0^{\rm{\pi }} {\rm{x}}.{\rm{cos}}2{\rm{x}}$.dx
Or, $\frac{1}{2}\mathop \smallint \limits_0^{\rm{\pi }} {\rm{x}}.$dx = $\frac{1}{2}$.$\left[ {\frac{{{{\rm{x}}^2}}}{2}} \right]_0^{\rm{\pi }}$ = $\frac{1}{4}$π2.
Or, $\mathop \smallint \nolimits {\rm{x}}.{\rm{cos}}2{\rm{x}}$.dx = x$\mathop \smallint \nolimits {\rm{cos}}2{\rm{x}}$.dx – $\mathop \smallint \nolimits \{ \frac{{\rm{d}}}{{{\rm{dx}}}}\left( {\rm{x}} \right)\mathop \smallint \nolimits {\rm{cos}}2{\rm{x}}.{\rm{dx}}\} $.dx
= x.$\frac{{{\rm{sin}}2{\rm{x}}}}{2}$ – $\mathop \smallint \nolimits 1.\frac{{{\rm{sin}}2{\rm{x}}}}{2}$.dx = $\frac{1}{2}$.x.sin2x – $\frac{1}{2}\left( { - \frac{{{\rm{cos}}2{\rm{x}}}}{2}} \right)$.
= $\frac{1}{2}$x.sin2x + $\frac{1}{4}$cos2x.
Or, $\mathop \smallint \limits_0^{\rm{\pi }} {\rm{x}}.{\rm{cos}}2{\rm{x}}$ = $\left[ {\frac{1}{2}{\rm{x}}.{\rm{sin}}2{\rm{x}} + \frac{1}{4}{\rm{cos}}2{\rm{x}}} \right]_0^{\rm{\pi }}$
=$\left( {\frac{1}{2}{\rm{\pi }}.{\rm{sin}}2{\rm{\pi }} + \frac{1}{4}{\rm{cos}}2{\rm{\pi }}} \right) - \left( {0 + \frac{1}{4}.{\rm{cos}}0} \right)$ = 0 + $\frac{1}{4}$.1 – $\frac{1}{4}$.1 = 0
So, $\mathop \smallint \limits_0^{\rm{\pi }} {\rm{x}}.{\sin ^2}{\rm{x}}$.dx = $\frac{1}{2}\mathop \smallint \limits_0^{\rm{\pi }} {\rm{x}}.{\rm{dx}} - \frac{1}{2}\mathop \smallint \limits_0^{\rm{\pi }} {\rm{x}}.{\rm{cos}}2{\rm{x}}$dx
= $\frac{1}{4}$π2 – $\frac{1}{2}$.0 = $\frac{{{{\rm{\pi }}^2}}}{4}$.
34. $\mathop \smallint \limits_0^{\rm{\pi }} .{\rm{x}}.{\cos ^2}{\rm{x}}$.dx
Solution:
Or, $\mathop \smallint \limits_0^{\rm{\pi }} .{\rm{x}}.{\cos ^2}{\rm{x}}$.dx= $\frac{1}{2}\mathop \smallint \limits_0^{\rm{\pi }} {\rm{x}}(2{\cos ^2}{\rm{x}}).{\rm{dx}}$ = $\frac{1}{2}$$\mathop \smallint \limits_0^{\rm{\pi }} \left( {{\rm{x}} + {\rm{xcos}}2{\rm{x}}} \right)$.dx
= $\frac{1}{2}\mathop \smallint \limits_0^{\rm{\pi }} {\rm{x}}.$dx + $\frac{1}{2}$$\mathop \smallint \limits_0^{\rm{\pi }} {\rm{x}}.{\rm{cos}}2{\rm{x}}$.dx
Or, $\frac{1}{2}\mathop \smallint \limits_0^{\rm{\pi }} {\rm{x}}.$dx = $\frac{1}{2}$.$\left[ {\frac{{{{\rm{x}}^2}}}{2}} \right]_0^{\rm{\pi }}$ = $\frac{1}{4}$π2.
Or, $\mathop \smallint \nolimits {\rm{x}}.{\rm{cos}}2{\rm{x}}$.dx = x$\mathop \smallint \nolimits {\rm{cos}}2{\rm{x}}$.dx – $\mathop \smallint \nolimits \{ \frac{{\rm{d}}}{{{\rm{dx}}}}\left( {\rm{x}} \right)\mathop \smallint \nolimits {\rm{cos}}2{\rm{x}}.{\rm{dx}}\} $.dx
= x.$\frac{{{\rm{sin}}2{\rm{x}}}}{2}$ – $\mathop \smallint \nolimits 1.\frac{{{\rm{sin}}2{\rm{x}}}}{2}$.dx = $\frac{1}{2}$.x.sin2x – $\frac{1}{2}\left( { - \frac{{{\rm{cos}}2{\rm{x}}}}{2}} \right)$.
= $\frac{1}{2}$x.sin2x + $\frac{1}{4}$cos2x.
Or, $\mathop \smallint \limits_0^{\rm{\pi }} {\rm{x}}.{\rm{cos}}2{\rm{x}}$ = $\left[ {\frac{1}{2}{\rm{x}}.{\rm{sin}}2{\rm{x}} + \frac{1}{4}{\rm{cos}}2{\rm{x}}} \right]_0^{\rm{\pi }}$
=$\left( {\frac{1}{2}{\rm{\pi }}.{\rm{sin}}2{\rm{\pi }} + \frac{1}{4}{\rm{cos}}2{\rm{\pi }}} \right) - \left( {0 + \frac{1}{4}.{\rm{cos}}0} \right)$ = 0 + $\frac{1}{4}$.1 – $\frac{1}{4}$.1 = 0
So, $\mathop \smallint \limits_0^{\rm{\pi }} {\rm{x}}.{\cos ^2}{\rm{x}}$.dx = $\frac{1}{2}\mathop \smallint \limits_0^{\rm{\pi }} {\rm{x}}.{\rm{dx}} + \frac{1}{2}\mathop \smallint \limits_0^{\rm{\pi }} {\rm{x}}.{\rm{cos}}2{\rm{x}}.$dx
= $\frac{1}{4}$π2 + $\frac{1}{2}$.0 = $\frac{{{{\rm{\pi }}^2}}}{4}$.
35. $\mathop \smallint \limits_0^{\rm{a}} \sqrt {{{\rm{a}}^2} - {{\rm{x}}^2}} $.dx
Solution:
Let I = $\mathop \smallint \limits_0^{\rm{a}} \sqrt {{{\rm{a}}^2} - {{\rm{x}}^2}} $.dx
Put x = a.sinθ
dx = a.cosθ.θ
When x = a, θ = $\frac{{\rm{\pi }}}{2}$; x = 0,θ = 0.
I = $\mathop \smallint \limits_0^{\frac{{\rm{\pi }}}{2}} \sqrt {{{\rm{a}}^2} - {{\rm{a}}^2}.{{\sin }^2}\theta } $.a.cosθ.dθ
= $\mathop \smallint \limits_0^{\frac{{\rm{\pi }}}{2}} {\rm{a}}.{\rm{cos}}\theta .{\rm{a}}.{\rm{cos}}\theta .$dθ = $\frac{{{{\rm{a}}^2}}}{2}\mathop \smallint \limits_0^{\frac{{\rm{\pi }}}{2}} 2{\cos ^2}\theta $.dθ
= $\frac{{{{\rm{a}}^2}}}{2}$$\mathop \smallint \limits_0^{\frac{{\rm{\pi }}}{2}} \left( {1 + {\rm{cos}}2\theta } \right)$.dθ
= $\frac{{{{\rm{a}}^2}}}{2}$$\left[ {\theta + \frac{{{\rm{sin}}2\theta }}{2}} \right]_0^{\frac{{\rm{\pi }}}{2}}$
= $\frac{{{{\rm{a}}^2}}}{2}$$\left[ {\frac{{\rm{\pi }}}{2} + \frac{{{\rm{sin\pi }}}}{2} - 0 - 0} \right]$ = $\frac{{{{\rm{a}}^2}}}{2}.\frac{{\rm{\pi }}}{2}$ = $\frac{{{{\rm{a}}^2}{\rm{\pi }}}}{4}$.
36. $\mathop \smallint \limits_0^{\frac{1}{2}} \frac{{{\rm{dx}}}}{{\sqrt {1 - {{\rm{x}}^2}} }}$
Solution:
Let I = $\mathop \smallint \limits_0^{\frac{1}{2}} \frac{{{\rm{dx}}}}{{\sqrt {1 - {{\rm{x}}^2}} }}$
Put x = sinθ ⇒ dx = cosθ.dθ
When x = 0, θ = 0; x = $\frac{1}{2}$, θ = $\frac{{\rm{\pi }}}{6}$.
I = $\mathop \smallint \limits_0^{\frac{{\rm{\pi }}}{6}} \frac{{{\rm{cos}}\theta .{\rm{d}}\theta }}{{\sqrt {1 - {{\sin }^2}\theta } }}{\rm{\: \: }}$
= $\mathop \smallint \limits_0^{\frac{{\rm{\pi }}}{6}} \frac{{{\rm{cos}}\theta .{\rm{d}}\theta }}{{\sqrt {{{\cos }^2}\theta } }}{\rm{\: }}$
= $\mathop \smallint \limits_0^{\frac{{\rm{\pi }}}{6}} \frac{{{\rm{cos}}\theta .{\rm{d}}\theta }}{{{\rm{cos}}\theta }}{\rm{\: }}$
= $\mathop \smallint \limits_0^{\frac{{\rm{\pi }}}{6}} {\rm{d}}\theta {\rm{\: }}$ = $\left[ \theta \right]_0^{\frac{{\rm{\pi }}}{6}}$=$\frac{{\rm{\pi }}}{6}$.
37. $\mathop \smallint \limits_0^1 \frac{{{\rm{dx}}}}{{1 + {{\rm{x}}^2}}}$
Solution:
Let I = $\mathop \smallint \limits_0^1 \frac{{{\rm{dx}}}}{{1 + {{\rm{x}}^2}}}$
Put x = tanθ ⇒ dx = sec2θ.dθ
When x = 0, θ = 0; x = 1, θ = $\frac{{\rm{\pi }}}{4}$.
I = $\mathop \smallint \limits_0^{\frac{{\rm{\pi }}}{4}} \frac{{{{\sec }^2}\theta .{\rm{d}}\theta }}{{1 + {{\tan }^2}\theta }}{\rm{\: \: }}$
= $\mathop \smallint \limits_0^{\frac{{\rm{\pi }}}{4}} \frac{{{{\sec }^2}\theta .{\rm{d}}\theta }}{{1 + {{\tan }^2}\theta }}{\rm{\: }}$
= $\mathop \smallint \limits_0^{\frac{{\rm{\pi }}}{4}} \frac{{{{\sec }^2}\theta .{\rm{d}}\theta }}{{{{\sec }^2}\theta }}{\rm{\: }}$
= $\mathop \smallint \limits_0^{\frac{{\rm{\pi }}}{4}} {\rm{d}}\theta {\rm{\: }}$ = $\left[ \theta \right]_0^{\frac{{\rm{\pi }}}{4}}$=$\frac{{\rm{\pi }}}{4}$.
38. $\mathop \smallint \limits_0^{\frac{{\rm{\pi }}}{2}} \frac{{{\rm{cosx}}}}{{1 + {{\sin }^2}{\rm{x}}}}$.dx
Solution:
Let I = $\mathop \smallint \limits_0^{\frac{{\rm{\pi }}}{2}} \frac{{{\rm{cosx}}}}{{1 + {{\sin }^2}{\rm{x}}}}$.dx
Put x = tanθ
Or, cosx.dx = sec2θ.dθ
When x = 0, θ = 0; x = $\frac{{\rm{\pi }}}{2}$, θ = $\frac{{\rm{\pi }}}{4}$.
I = $\mathop \smallint \limits_0^{\frac{{\rm{\pi }}}{4}} \frac{{{{\sec }^2}\theta .{\rm{d}}\theta }}{{1 + {{\tan }^2}\theta }}{\rm{\: \: }}$
= $\mathop \smallint \limits_0^{\frac{{\rm{\pi }}}{4}} \frac{{{{\sec }^2}\theta .{\rm{d}}\theta }}{{1 + {{\tan }^2}\theta }}{\rm{\: }}$
= $\mathop \smallint \limits_0^{\frac{{\rm{\pi }}}{4}} \frac{{{{\sec }^2}\theta .{\rm{d}}\theta }}{{{{\sec }^2}\theta }}{\rm{\: }}$
= $\mathop \smallint \limits_0^{\frac{{\rm{\pi }}}{4}} {\rm{d}}\theta {\rm{\: }}$ = $\left[ \theta \right]_0^{\frac{{\rm{\pi }}}{4}}$=$\frac{{\rm{\pi }}}{4}$.
39. $\mathop \smallint \limits_0^{ - 1} \frac{{{\rm{dx}}}}{{\sqrt {4 - {{\rm{x}}^2}} }}$
Solution:
Let I = $\mathop \smallint \limits_0^{ - 1} \frac{{{\rm{dx}}}}{{\sqrt {4 - {{\rm{x}}^2}} }}$
Put x = 2.sinθ ⇒ dx = 2cosθ.dθ
When x = 0,θ = 0; x =–1, θ = $ - \frac{{\rm{\pi }}}{6}$.
I = $\mathop \smallint \limits_0^{ - \frac{{\rm{\pi }}}{6}} \frac{{2{\rm{cos}}\theta .{\rm{d}}\theta }}{{\sqrt {4 - 4{{\sin }^2}\theta } }}{\rm{\: \: }}$
= $\mathop \smallint \limits_0^{ - \frac{{\rm{\pi }}}{6}} \frac{{2{\rm{cos}}\theta .{\rm{d}}\theta }}{{\sqrt {4\left( {1 - {{\sin }^2}\theta } \right)} }}{\rm{\: }}$
= $\mathop \smallint \limits_0^{ - \frac{{\rm{\pi }}}{6}} \frac{{2{\rm{cos}}\theta .{\rm{d}}\theta }}{{\sqrt {4{{\cos }^2}\theta } }}{\rm{\: }}$
= $\mathop \smallint \limits_0^{ - \frac{{\rm{\pi }}}{6}} \frac{{2{\rm{cos}}\theta .{\rm{d}}\theta }}{{2.{\rm{cos}}\theta }}{\rm{\: }}$
= $\mathop \smallint \limits_0^{ - \frac{{\rm{\pi }}}{6}} {\rm{d}}\theta {\rm{\: }}$ = $\left[ \theta \right]_0^{ - \frac{{\rm{\pi }}}{6}}$=$ - \frac{{\rm{\pi }}}{6}$.
40. $\mathop \smallint \limits_0^{\rm{a}} \frac{{{\rm{dx}}}}{{{{\rm{a}}^2} + {{\rm{x}}^2}}}$
Solution:
Let I = $\mathop \smallint \limits_0^{\rm{a}} \frac{{{\rm{dx}}}}{{{{\rm{a}}^2} + {{\rm{x}}^2}}}$
Put x = a.tanθ ⇒ dx = a.sec2θ.dθ
When x = 0,θ = 0; x = a, θ = $\frac{{\rm{\pi }}}{4}$.
I = $\mathop \smallint \limits_0^{\frac{{\rm{\pi }}}{4}} \frac{{{\rm{a}}.{{\sec }^2}\theta .{\rm{d}}\theta }}{{{{\rm{a}}^2} + {{\rm{a}}^2}{{\tan }^2}\theta }}{\rm{\: \: }}$
= $\mathop \smallint \limits_0^{\frac{{\rm{\pi }}}{4}} \frac{{{\rm{a}}.{{\sec }^2}\theta .{\rm{d}}\theta }}{{{{\rm{a}}^2}\left( {1 + {{\tan }^2}\theta } \right)}}{\rm{\: }}$
= $\frac{1}{{\rm{a}}}\mathop \smallint \limits_0^{\frac{{\rm{\pi }}}{4}} \frac{{{{\sec }^2}\theta .{\rm{d}}\theta }}{{{{\sec }^2}\theta }}{\rm{\: }}$
= $\frac{1}{{\rm{a}}}\mathop \smallint \limits_0^{\frac{{\rm{\pi }}}{4}} {\rm{d}}\theta {\rm{\: }}$
= $\frac{1}{{\rm{a}}}\mathop \smallint \limits_0^{\frac{{\rm{\pi }}}{4}} {\rm{d}}\theta {\rm{\: }}$ = $\frac{1}{{\rm{a}}}.\left[ \theta \right]_0^{\frac{{\rm{\pi }}}{4}}$=$\frac{1}{{\rm{a}}}.\frac{{\rm{\pi }}}{4}$ = $\frac{{\rm{\pi }}}{{4{\rm{a}}}}$