Exercise 19.3
Find the indefinite integral of
1.(i) $\mathop \smallint \nolimits \frac{{{\rm{dx}}}}{{\sqrt {1 - {{\rm{x}}^2}} }}$
Solution:
Let I = $\mathop \smallint \nolimits \frac{{{\rm{dx}}}}{{\sqrt {1 - {{\rm{x}}^2}} }}$
Put x = sinθ.
dx = cosθ.dθ
I = $\mathop \smallint \nolimits \frac{{{\rm{cos}}\theta .{\rm{d}}\theta }}{{\sqrt {1 - {{\sin }^2}\theta } }}$ = $\mathop \smallint \nolimits \frac{{{\rm{cos}}\theta }}{{{\rm{cos}}\theta }}$.dθ = $\mathop \smallint \nolimits {\rm{d}}\theta $ = θ + c.
So, I = sin–1x + c.
(ii) $\mathop \smallint \nolimits \frac{{{\rm{dx}}}}{{{{\rm{x}}^2}\sqrt {9 - {{\rm{x}}^2}} }}$
Solution:
Let I = $\mathop \smallint \nolimits \frac{{{\rm{dx}}}}{{{{\rm{x}}^2}\sqrt {9 - {{\rm{x}}^2}} }}$
Put x = 3sinθ ⇒ 3cosθ.dθ
I = $\mathop \smallint \nolimits \frac{{3.{\rm{cos}}\theta .{\rm{d}}\theta }}{{9{{\sin }^2}\theta \sqrt {9 - 9{{\sin }^2}\theta } }}$ = $\mathop \smallint \nolimits \frac{{3{\rm{cos}}\theta .{\rm{d}}\theta }}{{9{{\sin }^2}\theta \sqrt {9\left( {1 - {{\sin }^2}\theta } \right)} }}$
= $\mathop \smallint \nolimits \frac{{3{\rm{cos}}\theta .{\rm{d}}\theta }}{{9{{\sin }^2}\theta \sqrt {9{{\cos }^2}\theta } }}$
= $\mathop \smallint \nolimits \frac{{3{\rm{cos}}\theta .{\rm{d}}\theta }}{{9{{\sin }^2}\theta .3{\rm{cos}}\theta {\rm{\: }}}}$
= $\frac{1}{9}$$\mathop \smallint \nolimits {\rm{cose}}{{\rm{c}}^2}\theta .{\rm{d}}\theta $ = $\frac{1}{9}$(–cotθ) + c
= $ - \frac{1}{9}\frac{{{\rm{cos}}\theta }}{{{\rm{sin}}\theta }}{\rm{\: }}$ + c
= $ - \frac{1}{9}$$\frac{{\sqrt {1 - {{\sin }^2}\theta } }}{{{\rm{sin}}\theta }}$ + c.
= $ - \frac{1}{9}$$\frac{{\sqrt {1 - \frac{{{{\rm{x}}^2}}}{9}} }}{{\frac{{\rm{x}}}{3}}}{\rm{\: }}$ + c (x = 3sinθ)
= $ - \frac{{\sqrt {9 - {{\rm{x}}^2}} }}{{9{\rm{x}}}}$ + c.
(iii) $\mathop \smallint \nolimits \frac{{{\rm{dx}}}}{{{{\left( {{{\rm{a}}^2} - {{\rm{x}}^2}} \right)}^{\frac{3}{2}}}}}$
Solution:
Let I = $\mathop \smallint \nolimits \frac{{{\rm{dx}}}}{{{{\left( {{{\rm{a}}^2} - {{\rm{x}}^2}} \right)}^{\frac{3}{2}}}}}$
Put x = asinθ
dx = a.cosθ.dθ
I = $\mathop \smallint \nolimits \frac{{{\rm{a}}.{\rm{cos}}\theta .{\rm{d}}\theta }}{{{{\left( {{{\rm{a}}^2} - {{\rm{a}}^2}.{{\sin }^2}\theta } \right)}^{\frac{3}{2}}}}}$ = $\mathop \smallint \nolimits \frac{{{\rm{acos}}\theta .{\rm{d}}\theta }}{{{{\rm{a}}^3}.{{\cos }^3}\theta }}$
= $\frac{1}{{{{\rm{a}}^2}}}\mathop \smallint \nolimits {\sec ^2}\theta .{\rm{d}}\theta $ = $\frac{1}{{{{\rm{a}}^2}}}$tanθ + c = $\frac{1}{{{{\rm{a}}^2}}}$$\frac{{{\rm{sin}}\theta }}{{{\rm{cos}}\theta }}$ + c
= $\frac{1}{{{{\rm{a}}^2}}}.\frac{{{\rm{sin}}\theta }}{{\sqrt {1 - {{\sin }^2}\theta } }}$ + c = $\frac{1}{{{{\rm{a}}^2}}}$. $\frac{{\frac{{\rm{x}}}{{\rm{a}}}}}{{\sqrt {1 - \frac{{{{\rm{x}}^2}}}{{{{\rm{a}}^2}}}} }}$
= $\frac{{\rm{x}}}{{{{\rm{a}}^2}\left( {\sqrt {{{\rm{a}}^2} - {{\rm{x}}^2}} } \right)}}$ + c.
(iv) $\mathop \smallint \nolimits \frac{{{{\rm{x}}^2}.{\rm{dx}}}}{{\sqrt {{{\rm{a}}^2} - {{\rm{x}}^2}} }}$
Solution:
Let I = $\mathop \smallint \nolimits \frac{{{{\rm{x}}^2}.{\rm{dx}}}}{{\sqrt {{{\rm{a}}^2} - {{\rm{x}}^2}} }}$
Put x = asinθ
dx = a.cosθ.dθ
I = $\mathop \smallint \nolimits \frac{{{{\rm{a}}^2}.{{\sin }^2}\theta .{\rm{acos}}\theta .{\rm{d}}\theta }}{{\sqrt {{{\rm{a}}^2} - {{\rm{a}}^2}.{{\sin }^2}\theta } }}$ = $\mathop \smallint \nolimits \frac{{{{\rm{a}}^2}{{\sin }^2}\theta .{\rm{acos}}\theta .{\rm{d}}\theta }}{{\sqrt {{{\rm{a}}^2}\left( {1 - {{\sin }^2}\theta } \right)} }}$
= $\mathop \smallint \nolimits \frac{{{{\rm{a}}^2}.{{\sin }^2}\theta .{\rm{acos}}\theta .{\rm{d}}\theta }}{{\sqrt {{{\rm{a}}^2}.{{\cos }^2}\theta } }}$
=$\mathop \smallint \nolimits \frac{{{{\rm{a}}^2}.{{\sin }^2}\theta .{\rm{acos}}\theta .{\rm{d}}\theta }}{{{\rm{a}}.{\rm{cos}}\theta }}$
=$\frac{{{{\rm{a}}^2}}}{2}\mathop \smallint \nolimits 2{\sin ^2}\theta .{\rm{d}}\theta $ = $\frac{{{{\rm{a}}^2}}}{2}\mathop \smallint \nolimits \left( {1 - {\rm{cos}}2\theta } \right).{\rm{d}}\theta $
= $\frac{{{{\rm{a}}^2}}}{2}\left[ {\theta - \frac{{{\rm{sin}}2\theta }}{2}} \right]$ + c
= $\frac{{{{\rm{a}}^2}}}{2}\left[ {\theta - \frac{{2{\rm{sin}}\theta .{\rm{cos}}\theta }}{2}} \right]$ + c
= $\frac{{{{\rm{a}}^2}}}{2}$$\left[ {\theta - {\rm{sin}}\theta .{\rm{cos}}\theta } \right]$ + c
= $\frac{{{{\rm{a}}^2}}}{2}$$\left[ {\theta - {\rm{sin}}\theta \sqrt {1 - {{\sin }^2}\theta } } \right]$ + c
= $\frac{{{{\rm{a}}^2}}}{2}$$\left[ {{{\sin }^{ - 1}}\frac{{\rm{x}}}{{\rm{a}}} - \frac{{\rm{x}}}{{\rm{a}}}\sqrt {1 - \frac{{{{\rm{x}}^2}}}{{{{\rm{a}}^2}}}} } \right]$ + c
= $\frac{{{{\rm{a}}^2}}}{2}$sin–1$\frac{{\rm{x}}}{{\rm{a}}}$ – $\frac{1}{2}$ x$\sqrt {{{\rm{a}}^2} - {{\rm{x}}^2}} $ + c.
2.(i) $\mathop \smallint \nolimits \frac{{\sqrt {{{\rm{x}}^2} - 9} }}{{\rm{x}}}$.dx
Solution:
Let I = $\mathop \smallint \nolimits \frac{{\sqrt {{{\rm{x}}^2} - 9} }}{{\rm{x}}}$.dx
Put x = 3secθ ⇒ dx = 3secθ.tanθ.dθ
I = $\mathop \smallint \nolimits \frac{{\sqrt {9{{\sec }^2}\theta - 9} }}{{3{\rm{sec}}\theta }}$.3secθ.tanθ.dθ.
= $\mathop \smallint \nolimits \sqrt {9\left( {{{\sec }^2}\theta - 1} \right)} $.tanθ.dθ
= $\mathop \smallint \nolimits \sqrt {9{{\tan }^2}\theta } $.tanθ.dθ
= 3.$\mathop \smallint \nolimits {\tan ^2}\theta .{\rm{d}}\theta $ = 3$\mathop \smallint \nolimits \left( {{{\sec }^2}\theta - 1} \right).{\rm{d}}\theta $ = 3(tanθ – θ) + c
= 3.$\left[ {\frac{{\sqrt {{{\rm{x}}^2} - 9} }}{3} - {{\sec }^{ - 1}}\frac{{\rm{x}}}{3}} \right]$ + c = $\sqrt {{{\rm{x}}^2} - 9} $ – 3sec–1$\frac{{\rm{x}}}{3}$ + c.
(ii) $\mathop \smallint \nolimits \frac{{{\rm{dx}}}}{{{{\rm{x}}^2}.\sqrt {{{\rm{x}}^2} - {{\rm{a}}^2}} }}$.dx
Solution:
Let I = $\mathop \smallint \nolimits \frac{{{\rm{dx}}}}{{{{\rm{x}}^2}.\sqrt {{{\rm{x}}^2} - {{\rm{a}}^2}} }}$.dx
Put x = a.secθ ⇒ dx = a.secθ.tanθ.dθ
I = $\mathop \smallint \nolimits \frac{{{\rm{a}}.{\rm{sec}}\theta .{\rm{tan}}\theta .{\rm{d}}\theta }}{{{{\rm{a}}^2}.{{\sec }^2}\theta \sqrt {{{\rm{a}}^2}{{\sec }^2}\theta - {{\rm{a}}^2}} }}$
= $\mathop \smallint \nolimits \frac{{{\rm{a}}.{\rm{sec}}\theta .{\rm{tan}}\theta .{\rm{d}}\theta }}{{{{\rm{a}}^2}.{{\sec }^2}\theta \sqrt {{{\rm{a}}^2}({{\sec }^2}\theta - 1)} }}$
= $\mathop \smallint \nolimits \frac{{{\rm{a}}.{\rm{sec}}\theta .{\rm{tan}}\theta .{\rm{d}}\theta }}{{{{\rm{a}}^2}.{{\sec }^2}\theta \sqrt {{{\rm{a}}^2}.{{\tan }^2}\theta } }}$
= $\mathop \smallint \nolimits \frac{{{\rm{a}}.{\rm{sec}}\theta .{\rm{tan}}\theta .{\rm{d}}\theta }}{{{{\rm{a}}^2}.{{\sec }^2}\theta .{\rm{atan}}\theta }}$
= $\frac{1}{{{{\rm{a}}^2}}}\mathop \smallint \nolimits {\rm{cos}}\theta $.dθ
= $\frac{1}{{{{\rm{a}}^2}}}$.sinθ + c
= $\frac{1}{{{{\rm{a}}^2}}}$.$\frac{{\sqrt {{{\rm{x}}^2} - {{\rm{a}}^2}} }}{{\rm{x}}}$ + c
= $\frac{{\sqrt {{{\rm{x}}^2} - {{\rm{a}}^2}} }}{{{{\rm{a}}^2}{\rm{x}}}}$ + c.
(iii) $\mathop \smallint \nolimits \frac{{{\rm{dx}}}}{{{{\left( {{{\rm{x}}^2} - {{\rm{a}}^2}} \right)}^{\frac{3}{2}}}}}$.dx
Solution:
Let I = $\mathop \smallint \nolimits \frac{{{\rm{dx}}}}{{{{\left( {{{\rm{x}}^2} - {{\rm{a}}^2}} \right)}^{\frac{3}{2}}}}}$.dx
Put x = a.secθ ⇒ dx = a.secθ.tanθ.dθ
I = $\mathop \smallint \nolimits \frac{{{\rm{a}}.{\rm{sec}}\theta .{\rm{tan}}\theta .{\rm{d}}\theta }}{{{{\left( {{{\rm{a}}^2}.{{\sec }^2}\theta - {{\rm{a}}^2}} \right)}^{\frac{3}{2}}}}}$
= $\mathop \smallint \nolimits \frac{{{\rm{a}}.{\rm{sec}}\theta .{\rm{tan}}\theta .{\rm{d}}\theta }}{{{{\left\{ {{{\rm{a}}^2}\left( {{{\sec }^2}\theta - 1} \right)} \right\}}^{\frac{3}{2}}}}}$
= $\mathop \smallint \nolimits \frac{{{\rm{a}}.{\rm{sec}}\theta .{\rm{tan}}\theta .{\rm{d}}\theta }}{{{{\left( {{{\rm{a}}^2}.{{\tan }^2}\theta } \right)}^{\frac{3}{2}}}}}$
= $\mathop \smallint \nolimits \frac{{{\rm{a}}.{\rm{sec}}\theta .{\rm{tan}}\theta .{\rm{d}}\theta }}{{{{\rm{a}}^3}.{{\tan }^3}\theta }}$
= $\frac{1}{{{{\rm{a}}^2}}}\mathop \smallint \nolimits {\rm{cot}}\theta .{\rm{cosec}}\theta $.dθ
= $ - \frac{1}{{{{\rm{a}}^2}}}$.cosecθ + c
= $ - \frac{1}{{{{\rm{a}}^2}}}$.$\frac{{\rm{x}}}{{\sqrt {{{\rm{x}}^2} - {{\rm{a}}^2}} }}$ + c.
(iv) $\mathop \smallint \nolimits \frac{{{\rm{dx}}}}{{\sqrt {{{\rm{x}}^2} - 4} }}$.dx
Solution:
Let I = $\mathop \smallint \nolimits \frac{{{\rm{dx}}}}{{\sqrt {{{\rm{x}}^2} - 4} }}$.dx
Put x = 2.secθ
dx = 2.secθ.tanθ.dθ
I = $\mathop \smallint \nolimits \frac{{2.{\rm{sec}}\theta .{\rm{tan}}\theta .{\rm{d}}\theta }}{{\sqrt {4{{\sec }^2}\theta - 4} }}$
= $\mathop \smallint \nolimits \frac{{2.{\rm{sec}}\theta .{\rm{tan}}\theta .{\rm{d}}\theta }}{{2{\rm{tan}}\theta }}$
= $\mathop \smallint \nolimits {\rm{sec}}\theta $.dθ = log(secθ + tanθ) + c.
= log(secθ + $\sqrt {{{\sec }^2}\theta - 1} $) + c
= log$\left( {\frac{{\rm{x}}}{2} + \sqrt {\frac{{{{\rm{x}}^2}}}{4} - 1} } \right)$ + c
= log$\left[ {\frac{{{\rm{x}} + \sqrt {{{\rm{x}}^2} - 4} }}{2}} \right]$ + c
= log(x + $\sqrt {{{\rm{x}}^2} - 4} $) – log2 + c = log (x + $\sqrt {{{\rm{x}}^2} - 4} $) + c.
(v) $\mathop \smallint \nolimits \frac{{{{\rm{x}}^3}{\rm{dx}}}}{{{{\left( {{{\rm{x}}^2} - {{\rm{a}}^2}} \right)}^{\frac{3}{2}}}}}$.dx
Solution:
Let I = $\mathop \smallint \nolimits \frac{{{{\rm{x}}^3}{\rm{dx}}}}{{{{\left( {{{\rm{x}}^2} - {{\rm{a}}^2}} \right)}^{\frac{3}{2}}}}}$.dx
Put x = a.secθ
dx = a.secθ.tanθ.dθ
I = $\mathop \smallint \nolimits \frac{{{{\rm{a}}^3}.{{\sec }^3}\theta .{\rm{a}}.{\rm{sec}}\theta .{\rm{tan}}\theta .{\rm{d}}\theta }}{{{{\left( {{{\rm{a}}^2}{{\sec }^2}\theta - {{\rm{a}}^2}} \right)}^{\frac{3}{2}}}}}$
= $\mathop \smallint \nolimits \frac{{{{\rm{a}}^4}.{{\sec }^4}\theta .{\rm{tan}}\theta .{\rm{d}}\theta }}{{{{\left\{ {{{\rm{a}}^2}\left( {{{\sec }^2}\theta - 1} \right)} \right\}}^{\frac{3}{2}}}}}$
= $\mathop \smallint \nolimits \frac{{{{\rm{a}}^4}.{{\sec }^4}\theta .{\rm{tan}}\theta .{\rm{d}}\theta }}{{{{\rm{a}}^3}.{{\tan }^3}{\rm{\: }}\theta }}$
= ${\rm{a}}\mathop \smallint \nolimits \frac{{\frac{1}{{{{\cos }^4}\theta }}}}{{{{\sin }^2}\theta .{{\cos }^2}\theta }}$.dθ
= ${\rm{a}}\mathop \smallint \nolimits \frac{1}{{{{\cos }^4}\theta }}.\frac{{{{\cos }^2}\theta }}{{{{\sin }^2}\theta }}$.dθ
= ${\rm{a}}\mathop \smallint \nolimits \frac{1}{{{{\sin }^2}\theta .{{\cos }^2}\theta }}$.dθ
= ${\rm{a}}\mathop \smallint \nolimits \frac{{{{\sin }^2}\theta . + {{\cos }^2}\theta }}{{{{\sin }^2}\theta .{{\cos }^2}\theta }}$.dθ
= ${\rm{a}}\mathop \smallint \nolimits [\frac{{{{\sin }^2}\theta }}{{{{\sin }^2}\theta .{{\cos }^2}\theta }} + \frac{{{{\cos }^2}\theta }}{{{{\sin }^2}\theta .{{\cos }^2}\theta }}].{\rm{d}}\theta $
= a$\mathop \smallint \nolimits ({\sec ^2}\theta + {\rm{cose}}{{\rm{c}}^2}\theta )$.dθ = a[tanθ – cotθ] + c
= ${\rm{a}}\left[ {\frac{{\alpha \left( {\sqrt {{{\rm{x}}^2} - {{\rm{a}}^2}} } \right)}}{{\rm{a}}} - \frac{{\rm{a}}}{{\sqrt {{{\rm{x}}^2} - {{\rm{a}}^2}} }}} \right]$ + c
= a. $\frac{{{{\rm{x}}^2} - {{\rm{a}}^2} - {{\rm{a}}^2}}}{{{\rm{a}}\sqrt {{{\rm{x}}^2} - {{\rm{a}}^2}} }}$ = $\frac{{{{\rm{x}}^2} - 2{{\rm{a}}^2}}}{{\sqrt {{{\rm{x}}^2} - {{\rm{a}}^2}} }}$.
3.(i) $\mathop \smallint \nolimits \frac{{{\rm{dx}}}}{{\left( {{{\rm{x}}^2} + {{\rm{a}}^2}} \right)}}$
Solution:
Let I = $\mathop \smallint \nolimits \frac{{{\rm{dx}}}}{{\left( {{{\rm{x}}^2} + {{\rm{a}}^2}} \right)}}$
Put x = atanθ.
dx = asec2θ.dθ
I = $\mathop \smallint \nolimits \frac{{{\rm{a}}.{{\sec }^2}\theta .{\rm{d}}\theta }}{{{{\rm{a}}^2}{{\tan }^2}\theta + {{\rm{a}}^2}}}$
= $\frac{1}{{\rm{a}}}\mathop \smallint \nolimits {\rm{d}}\theta \frac{{\theta + {\rm{c}}}}{{\rm{a}}}$
= $\frac{1}{{\rm{a}}}$tan–1$\frac{{\rm{x}}}{{\rm{a}}}$ + c.
(ii) $\mathop \smallint \nolimits \frac{{{\rm{dx}}}}{{{{\rm{x}}^2}\sqrt {{{\rm{x}}^2} + 1} }}$
Solution:
Let I = $\mathop \smallint \nolimits \frac{{{\rm{dx}}}}{{{{\rm{x}}^2}\sqrt {{{\rm{x}}^2} + 1} }}$
Put x = tanθ ⇒ dx = sec2θ.dθ
dx = asec2θ.dθ
I = $\mathop \smallint \nolimits \frac{{.{{\sec }^2}\theta .{\rm{d}}\theta }}{{{{\tan }^2}\theta \sqrt {1 + {{\tan }^2}\theta } }}$
= $\mathop \smallint \nolimits \frac{{.{{\sec }^2}\theta .{\rm{d}}\theta }}{{{{\tan }^2}\theta {\rm{sec}}\theta }}$
= $\mathop \smallint \nolimits \frac{{{\rm{sec}}\theta }}{{{{\tan }^2}\theta }}$.dθ
= $\mathop \smallint \nolimits {\rm{cosec}}\theta .{\rm{cot}}\theta .{\rm{d}}\theta $ = –cosecθ + c
= $\frac{{ - \sqrt {{{\rm{x}}^2} + 1} }}{{\rm{x}}}$ + c
(iii) $\mathop \smallint \nolimits \frac{{{{\rm{x}}^2}.{\rm{d}}\theta }}{{{{\tan }^2}\theta \sqrt {1 + {{\tan }^2}\theta } }}$
Solution:
Let I = $\mathop \smallint \nolimits \frac{{{{\rm{x}}^2}.{\rm{d}}\theta }}{{{{\tan }^2}\theta \sqrt {1 + {{\tan }^2}\theta } }}$
Put x = tanθ ⇒ dx = sec2θ.dθ
I = $\mathop \smallint \nolimits \frac{{{\rm{ta}}{{\rm{n}}^2}.\theta .{{\sec }^2}\theta .{\rm{d}}\theta }}{{{{\left( {1 + {{\tan }^2}\theta } \right)}^2}}}$= $\mathop \smallint \nolimits \frac{{{{\tan }^2}\theta .{\rm{d}}\theta }}{{{{\sec }^2}\theta }}$
= $\mathop \smallint \nolimits {\sin ^2}\theta $.dθ
= $\frac{1}{2}\mathop \smallint \nolimits 2{\sin ^2}\theta $.dθ = $\frac{1}{2}\mathop \smallint \nolimits \left( {1 - {\rm{cos}}2\theta } \right)$.dθ
= $\frac{1}{2}\left[ {\theta - \frac{{{\rm{sin}}2\theta }}{2}} \right]$ = $\frac{1}{2}\left( {\theta - \frac{1}{2}{\rm{sin}}2\theta = \frac{1}{2}\left( {\theta - {\rm{sin}}\theta .{\rm{cos}}\theta } \right)} \right)$
= $\frac{1}{2}\left[ {{{\tan }^{ - 1}}{\rm{x}} - \frac{{\rm{x}}}{{\sqrt {1 + {{\rm{x}}^2}} }}.\frac{1}{{\sqrt {1 + {{\rm{x}}^2}} }}} \right]$ = $\frac{1}{2}$tan–1x $ - \frac{1}{2}$.$\frac{{\rm{x}}}{{1 + {{\rm{x}}^2}}}$ + c.
(iv) $\mathop \smallint \nolimits \frac{{{\rm{dx}}}}{{\sqrt {{{\rm{a}}^2} + {{\rm{x}}^2}} }}$
Solution:
Let I = $\mathop \smallint \nolimits \frac{{{\rm{dx}}}}{{\sqrt {{{\rm{a}}^2} + {{\rm{x}}^2}} }}$
Put x = a.tanθ ⇒ dx = a.sec2θ.dθ
I = $\mathop \smallint \nolimits \frac{{{\rm{a}}.{\rm{se}}{{\rm{c}}^2}\theta .{\rm{d}}\theta }}{{\sqrt {{\rm{a}} + {{\rm{a}}^2}{{\tan }^2}\theta } }}$= $\mathop \smallint \nolimits \frac{{{\rm{a}}.{{\sec }^2}\theta }}{{{\rm{a}}.{\rm{sec}}\theta }}$.dθ = $\mathop \smallint \nolimits {\rm{sec}}\theta .$dθ = log(secθ + tanθ) + c
= log$\left( {\frac{{\sqrt {{{\rm{x}}^2} + {{\rm{a}}^2}} }}{{\rm{a}}} + \frac{{\rm{x}}}{{\rm{a}}}} \right)$ + c = log$\left( {\frac{{{\rm{x}} + \sqrt {{{\rm{x}}^2} + {{\rm{a}}^2}} }}{{\rm{a}}}} \right)$ + c.
= log(x + $\sqrt {{{\rm{x}}^2} + {{\rm{a}}^2}} $) – loga + c = log(x + $\sqrt {{{\rm{x}}^2} + {{\rm{a}}^2}} $) + c.
(v) $\mathop \smallint \nolimits \frac{{{\rm{dx}}}}{{{\rm{x}}\sqrt {{{\rm{x}}^2} + 1} }}$
Solution:
Let I = $\mathop \smallint \nolimits \frac{{{\rm{dx}}}}{{{\rm{x}}\sqrt {{{\rm{x}}^2} + 1} }}$
Put x = tanθ ⇒ dx = sec2θ.dθ
Then $\sqrt {{{\rm{x}}^2} + 1} $ = $\sqrt {{{\tan }^2}\theta + 1} $ = secθ.
I = $\mathop \smallint \nolimits \frac{{{\rm{dx}}}}{{{\rm{x}}\sqrt {{{\rm{x}}^2} + 1{\rm{\: }}} }}$= $\mathop \smallint \nolimits \frac{{{{\sec }^2}\theta .{\rm{d}}\theta }}{{{\rm{tan}}\theta .{\rm{sec}}\theta }}$. = $\mathop \smallint \nolimits \frac{1}{{{\rm{sin}}\theta }}.$dθ = $\mathop \smallint \nolimits {\rm{cosec}}\theta $.dθ
= log(cosecθ – cotθ) + c
= log$\left( {\frac{{\sqrt {{{\rm{x}}^2} + 1} }}{{\rm{x}}} - \frac{1}{{\rm{x}}}} \right)$ + c = log $\frac{{\sqrt {{{\rm{x}}^2} + 1} - 1}}{{\rm{x}}}$ + c.
(vi) \[\int {\frac{{{{\tan }^{ - 1}}x}}{{1 + {x^2}}}dx} \]
Solution:
Put y = tan–1x
Or, $\frac{{{\rm{dy}}}}{{{\rm{dx}}}}$ = $\frac{1}{{1 + {{\rm{x}}^2}}}$
Or, dy = $\frac{1}{{{{\left( {1 + {\rm{x}}} \right)}^2}}}$.dx
Above integral reduced to:
Or, $\mathop \smallint \nolimits {\rm{y}}$.dy = $\frac{{{{\rm{y}}^2}}}{2}$ + c
Putting the value if y,we get,
So, $\frac{{{\rm{dy}}}}{{{\rm{dx}}}}$ = $\frac{{{{\left( {{{\tan }^{ - 1}}{\rm{x}}} \right)}^2}}}{2}$ + c.
4.(i) $\mathop \smallint \nolimits \sqrt {\frac{{{\rm{a}} + {\rm{x}}}}{{{\rm{a}} - {\rm{x}}}}} $.dx
Solution:
Let I = $\mathop \smallint \nolimits \sqrt {\frac{{{\rm{a}} + {\rm{x}}}}{{{\rm{a}} - {\rm{x}}}}} $.dx = $\mathop \smallint \nolimits \sqrt {\frac{{{\rm{a}} + {\rm{x}}}}{{{\rm{a}} - {\rm{x}}}}{\rm{*}}\frac{{{\rm{a}} + {\rm{x}}}}{{{\rm{a}} + {\rm{x}}}}} $ .dx = $\mathop \smallint \nolimits \frac{{{\rm{a}} + {\rm{x}}}}{{\sqrt {{{\rm{a}}^2} - {{\rm{x}}^2}} }}$.dx
Put x = acosθ ⇒ dx = – asinθ.dθ
I = $\mathop \smallint \nolimits \frac{{{\rm{a}} + {\rm{acos}}\theta }}{{\sqrt {{{\rm{a}}^2} - {{\rm{a}}^2}.{{\cos }^2}\theta } }}$.(–asinθ).dθ
= $ - \mathop \smallint \nolimits \frac{{{\rm{a}}\left( {1 + {\rm{cos}}\theta } \right)}}{{\sqrt {{{\rm{a}}^2}\left( {1 - {{\cos }^2}\theta } \right)} }}$.asinθdθ
= $ - \mathop \smallint \nolimits \frac{{{\rm{a}}\left( {1 + {\rm{cos}}\theta } \right)}}{{\sqrt {{{\rm{a}}^2}{{\sin }^2}\theta } }}$.asinθdθ
= $ - \mathop \smallint \nolimits \frac{{{\rm{a}}\left( {1 + {\rm{cos}}\theta } \right).{\rm{asin}}\theta {\rm{d}}\theta }}{{{\rm{a}}.{\rm{sin}}\theta }}$ = $ - {\rm{a}}\mathop \smallint \nolimits \left( {1 + {\rm{cos}}\theta } \right)$.dθ
= –a[θ + sinθ] + c = –a $\left[ {{{\cos }^{ - 1}}\frac{{\rm{x}}}{{\rm{a}}} + \sqrt {1 - {{\cos }^2}\theta } } \right]$ + c
= –a$\left[ {{{\cos }^{ - 1}}\frac{{\rm{x}}}{{\rm{a}}} + \sqrt {1 - \frac{{{{\rm{x}}^2}}}{{{{\rm{a}}^2}}}} } \right]$ + c (x = acosθ)
= –acos–1$\frac{{\rm{x}}}{{\rm{a}}}$–$\sqrt {{{\rm{a}}^2} - {{\rm{x}}^2}} $ + c.
(ii) $\mathop \smallint \nolimits \sqrt {\frac{{\rm{x}}}{{{\rm{a}} - {\rm{x}}}}} $.dx
Solution:
Let I = $\mathop \smallint \nolimits \sqrt {\frac{{\rm{x}}}{{{\rm{a}} - {\rm{x}}}}} $.dx
Put x = asin2θ ⇒ dx = a.2sinθ.cosθ.dθ
I = $\mathop \smallint \nolimits \sqrt {\frac{{{\rm{asi}}{{\rm{n}}^2}\theta }}{{{\rm{a}} - {\rm{asi}}{{\rm{n}}^2}\theta }}} $.2asinθ.cosθ.dθ
= $\sqrt {\frac{{{{\sin }^2}\theta }}{{{{\cos }^2}\theta }}} $.2asinθ.cosθ.dθ = $\mathop \smallint \nolimits \frac{{{\rm{sin}}\theta }}{{{\rm{cos}}\theta }}$ .2a.sinθ.cosθ.dθ
= a$\mathop \smallint \nolimits 2{\sin ^2}\theta .{\rm{d}}\theta $ = ${\rm{a}}\mathop \smallint \nolimits \left( {1 - {\rm{cos}}2\theta } \right)$.dθ
= a[θ – sinθ.cosθ] + c
Put x = a.sin2θ ⇒ sin2θ = $\frac{{\rm{x}}}{{\rm{a}}}$ So, sinθ= $\sqrt {\frac{{\rm{x}}}{{\rm{a}}}} $.
θ = sin–1$\sqrt {\frac{{\rm{x}}}{{\rm{a}}}} $ and cos θ = $\sqrt {1 - {{\sin }^2}\theta } $ = $\sqrt {1 - \frac{{\rm{x}}}{{\rm{a}}}} $
1 = a[θ – sinθ.cosθ] + c
= a$\left[ {{{\sin }^{ - 1}}\sqrt {\frac{{\rm{x}}}{{\rm{a}}}} - \sqrt {\frac{{\rm{x}}}{{\rm{a}}}} \sqrt {\frac{{{\rm{a}} - {\rm{x}}}}{{\rm{a}}}} } \right]$ + c = a.sin–1$\sqrt {\frac{{\rm{x}}}{{\rm{a}}}} $ – $\sqrt {{\rm{x}}\left( {{\rm{a}} - {\rm{x}}} \right)} $ + c