Anti-Derivatives Exercise: 19.4 Class 11 Basic Mathematics Solution [NEB UPDATED]

Exercise 19.4

1. Calculate the integrals:

a. $\mathop \smallint \nolimits {\rm{x}}.{\rm{logx}}$.dx

Solution:

Or, $\mathop \smallint \nolimits {\rm{x}}.{\rm{logx}}$.dx = logx $\mathop \smallint \nolimits {\rm{x}}$.dx – $\mathop \smallint \nolimits \{ \frac{{\rm{d}}}{{{\rm{dx}}}}\left( {{\rm{logx}}} \right)\mathop \smallint \nolimits {\rm{x}}.{\rm{dx}}\} {\rm{\: }}$dx.

= $\frac{{{{\rm{x}}^2}}}{2}$logx – $\mathop \smallint \nolimits \frac{1}{{\rm{x}}}.\frac{{{{\rm{x}}^2}}}{2}$.dx = $\frac{1}{2}$x²logx – $\frac{1}{2}\mathop \smallint \nolimits {\rm{x}}.$dx

= $\frac{1}{2}$x².logx – $\frac{1}{2}$.$\frac{{{{\rm{x}}^2}}}{2}$ + c = $\frac{1}{4}$x² (2logx – 1) + c.

 

b. ${\rm{\: }}\mathop \smallint \nolimits \left( {5{\rm{x}} - 2} \right).{\rm{logx}}$.dx

Solution:

Or, ${\rm{\: }}\mathop \smallint \nolimits \left( {5{\rm{x}} - 2} \right).{\rm{logx}}$.dx = logx $\mathop \smallint \nolimits \left( {5{\rm{x}} - 2} \right)$.dx – $\mathop \smallint \nolimits \{ \frac{{\rm{d}}}{{{\rm{dx}}}}\left( {{\rm{logx}}} \right)\mathop \smallint \nolimits \left( {5{\rm{x}} - 2} \right).{\rm{dx}}\} {\rm{\: }}$dx.

= log x $\left( {\frac{{5{{\rm{x}}^2}}}{2} - 2{\rm{x}}} \right)$ – $\mathop \smallint \nolimits \frac{1}{{\rm{x}}}$$\left( {\frac{{5{{\rm{x}}^2}}}{2} - 2{\rm{x}}} \right)$.dx

= $\left( {\frac{{5{{\rm{x}}^2}}}{2} - 2{\rm{x}}} \right)$.logx – $\mathop \smallint \nolimits \left( {\frac{{5{\rm{x}}}}{2} - 2} \right)$.dx

= $\frac{1}{2}$x(5x – 4).logx – $\left( {\frac{{5{{\rm{x}}^2}}}{{2{\rm{*}}2}} - 2{\rm{x}}} \right)$ + c.

= $\frac{1}{2}$x(5x – 4)logx – $\frac{5}{4}$ x² + 2x + c.

 

c. ${\rm{\: }}\mathop \smallint \nolimits {{\rm{x}}^{\rm{n}}}{\rm{logx}}$.dx

Solution:

Or, ${\rm{\: }}\mathop \smallint \nolimits {{\rm{x}}^{\rm{n}}}{\rm{logx}}$.dx = logx $\mathop \smallint \nolimits {{\rm{x}}^{\rm{n}}}$.dx – $\mathop \smallint \nolimits \{ \frac{{\rm{d}}}{{{\rm{dx}}}}\left( {{\rm{logx}}} \right)\mathop \smallint \nolimits {{\rm{x}}^{\rm{n}}}.{\rm{dx}}\} {\rm{\: }}$dx.

= log x $\frac{{{{\rm{x}}^{{\rm{n}} + 1}}}}{{{\rm{n}} + 1}}$ – $\mathop \smallint \nolimits \frac{1}{{\rm{x}}}\frac{{{{\rm{x}}^{{\rm{n}} + 1}}}}{{{\rm{n}} + 1}}$.dx = $\frac{{{{\rm{x}}^{{\rm{n}} + 1}}}}{{{\rm{n}} + 1}}$.logx – $\frac{1}{{{\rm{n}} + 1}}$$\mathop \smallint \nolimits {{\rm{x}}^{\rm{n}}}$.dx

= $\frac{{{{\rm{x}}^{{\rm{n}} + 1}}}}{{{\rm{n}} + 1}}$.logx – $\frac{1}{{{\rm{n}} + 1}}$.$\frac{{{{\rm{x}}^{{\rm{n}} + 1}}}}{{{\rm{n}} + 1}}$ + c

= $\frac{{{{\rm{x}}^{{\rm{n}} + 1}}}}{{{{\left( {{\rm{n}} + 1} \right)}^2}}}$[(n + 1)logx – 1] + c.

 

d. $\mathop \smallint \nolimits {\left( {\log {\rm{x}}} \right)^2}$.dx

Solution:

Or, $\mathop \smallint \nolimits {\left( {\log {\rm{x}}} \right)^2}$.dx= (logx)²$\mathop \smallint \nolimits 1$.dx – $\mathop \smallint \nolimits \{ \frac{{\rm{d}}}{{{\rm{dx}}}}{\left( {{\rm{logx}}} \right)^2}\mathop \smallint \nolimits 1.{\rm{dx}}\} $.dx

= (logx)²$\mathop \smallint \nolimits 1.{\rm{dx}}$– ≤$\mathop \smallint \nolimits \{ \frac{{{{\left( {{\rm{d}}\left( {{\rm{logx}}} \right)} \right)}^2}}}{{{\rm{d}}\left( {{\rm{logx}}} \right)}}.\frac{{{\rm{d}}\left( {{\rm{logx}}} \right)}}{{{\rm{dx}}}}\mathop \smallint \nolimits 1.{\rm{dx}}\} $.dx

= (logx)².x – $\mathop \smallint \nolimits 2.{\rm{log}}.{\rm{x}}.\frac{1}{{\rm{x}}}$.dx

= x(logx)² – 2$\mathop \smallint \nolimits 1.{\rm{logx}}$.dx

= x(logx)² – 2$\left[ {{\rm{logx}}\mathop \smallint \nolimits 1.{\rm{dx}} - \mathop \smallint \nolimits \{ \frac{{\rm{d}}}{{{\rm{dx}}}}\left( {{\rm{logx}}} \right)\mathop \smallint \nolimits 1.{\rm{dx}}\} .{\rm{dx}}} \right]$

= x(logx)² – 2[logx.x – $\mathop \smallint \nolimits \frac{1}{{\rm{x}}}$.x dx]

= x(logx)² – 2x.logx + 2x + c.

 

e. $\mathop \smallint \nolimits {\rm{xlog}}\left( {{\rm{x}} - 1} \right)$.dx

Solution:

Or, $\mathop \smallint \nolimits {\rm{xlog}}\left( {{\rm{x}} - 1} \right)$.dx= log(x – 1)$\mathop \smallint \nolimits {\rm{x}}$.dx – $\mathop \smallint \nolimits \{ \frac{{\rm{d}}}{{{\rm{dx}}}}{\rm{log}}\left( {{\rm{x}} - 1} \right)\mathop \smallint \nolimits {\rm{x}}.{\rm{dx}}\} $.dx

= log(x – 1) $\mathop \smallint \nolimits {\rm{x}}.{\rm{dx}}$– ≤$\mathop \smallint \nolimits \{ \frac{{{{\left( {{\rm{d}}\left( {{\rm{log}}\left( {{\rm{x}} - 1} \right)} \right)} \right)}^2}}}{{{\rm{d}}\left( {{\rm{log}}\left( {{\rm{x}} - 1} \right)} \right)}}.\frac{{{\rm{d}}\left( {{\rm{log}}\left( {{\rm{x}} - 1} \right)} \right)}}{{{\rm{dx}}}}\mathop \smallint \nolimits {\rm{x}}.{\rm{dx}}\} $.dx

= log(x – 1).$\frac{{{{\rm{x}}^2}}}{2}$ – $\mathop \smallint \nolimits \frac{1}{{{\rm{x}} - 1}}.1.\frac{{{{\rm{x}}^2}}}{2}$.dx

= $\frac{{{{\rm{x}}^2}}}{2}$log (x – 1) – $\frac{1}{2}$$\mathop \smallint \nolimits \frac{{\left( {{{\rm{x}}^2} - 1} \right) + 1}}{{{\rm{x}} - 1}}$.dx

= $\frac{{{{\rm{x}}^2}}}{2}$log(x – 1) – $\frac{1}{2}$$\mathop \smallint \nolimits \frac{{\left( {{\rm{x}} + 1} \right)\left( {{\rm{x}} - 1} \right) + 1}}{{{\rm{x}} - 1}}$.dx

= $\frac{{{{\rm{x}}^2}}}{2}$log(x – 1) – $\frac{1}{2}\mathop \smallint \nolimits \left( {{\rm{x}} + 1 + \frac{1}{{{\rm{x}} - 1}}} \right)$.dx

= $\frac{{{{\rm{x}}^2}}}{2}$ log(x – 1) – $\frac{1}{2}$$\left[ {\frac{{{{\rm{x}}^2}}}{2} + {\rm{x}} + \log \left( {{\rm{x}} - 1} \right)} \right]$ + c

= $\frac{{{{\rm{x}}^2}}}{2}$log(x – 1) – $\frac{1}{4}$x² – $\frac{1}{2}$x – $\frac{1}{2}$log(x – 1) + c

= $\frac{1}{2}$(x² – 1).log(x – 1) – $\frac{1}{4}$x² – $\frac{1}{2}$x + c.

 

f. $\mathop \smallint \nolimits {{\rm{x}}^2}.{{\rm{e}}^{{\rm{ax}}}}$.dx

Solution:

Or, $\mathop \smallint \nolimits {{\rm{x}}^2}.{{\rm{e}}^{{\rm{ax}}}}$.dx= x²$\mathop \smallint \nolimits {{\rm{e}}^{{\rm{ax}}}}$.dx – $\mathop \smallint \nolimits \{ \frac{{\rm{d}}}{{{\rm{dx}}}}\left( {{{\rm{x}}^2}} \right)\mathop \smallint \nolimits {{\rm{e}}^{{\rm{ax}}}}.{\rm{dx}}\} $.dx

= x². $\frac{{{{\rm{e}}^{{\rm{ax}}}}}}{{\rm{a}}}$–$\mathop \smallint \nolimits 2{\rm{x}}.\frac{{{{\rm{e}}^{{\rm{ax}}}}}}{{\rm{a}}}$.dx = $\frac{1}{{\rm{a}}}$x².e^ax – $\frac{2}{{\rm{a}}}\mathop \smallint \nolimits {\rm{x}}.{{\rm{e}}^{{\rm{ax}}}}$.dx

= $\frac{1}{{\rm{a}}}$x².e^ax – $\frac{2}{{\rm{a}}}[{\rm{x}}\mathop \smallint \nolimits {{\rm{e}}^{{\rm{ax}}}}.{\rm{dx}} - \mathop \smallint \nolimits \{ \frac{{\rm{d}}}{{{\rm{dx}}}}\left( {\rm{x}} \right)\mathop \smallint \nolimits {{\rm{e}}^{{\rm{ax}}}}.{\rm{dx}}\} .{\rm{dx}}]{\rm{\: }}$

= $\frac{1}{{\rm{a}}}$x².e^ax – $\frac{2}{{\rm{a}}}$$[{\rm{x}}.\frac{{{{\rm{e}}^{{\rm{ax}}}}}}{{\rm{a}}} - \mathop \smallint \nolimits {\rm{q}}.\frac{{{{\rm{e}}^{{\rm{ax}}}}}}{{\rm{a}}}.{\rm{dx}}]{\rm{\: }}$

= $\frac{1}{{\rm{a}}}$x²e^ax – $\frac{2}{{{{\rm{a}}^2}}}$x.e^ax + $\frac{2}{{\rm{a}}}$.$\frac{1}{{\rm{a}}}$.$\frac{{{{\rm{e}}^{{\rm{ax}}}}}}{{\rm{a}}}$ + c

= $\frac{1}{{\rm{a}}}$x².e^ax – $\frac{2}{{{{\rm{a}}^2}}}$x.e^ax + $\frac{2}{{{{\rm{a}}^3}}}$e^ax + c.

 

g. $\mathop \smallint \nolimits {\rm{x}}.{{\rm{e}}^{5{\rm{x}}}}$.dx

Solution:

Or, $\mathop \smallint \nolimits {\rm{x}}.{{\rm{e}}^{5{\rm{x}}}}$.dx= x$\mathop \smallint \nolimits {{\rm{e}}^{5{\rm{x}}}}$.dx – $\mathop \smallint \nolimits \{ \frac{{\rm{d}}}{{{\rm{dx}}}}\left( {\rm{x}} \right)\mathop \smallint \nolimits {{\rm{e}}^{5{\rm{x}}}}{\rm{dx}}\} {\rm{\: }}$.dx

= x.$\frac{{{{\rm{e}}^{5{\rm{x}}}}}}{5}$ – $\mathop \smallint \nolimits 1.\frac{{{{\rm{e}}^{5{\rm{x}}}}}}{5}.{\rm{dx\: }}$ = $\frac{1}{5}$x.e^5x – $\frac{1}{5}$. $\frac{{{{\rm{e}}^{5{\rm{x}}}}}}{5}$ + c

= $\frac{1}{{25}}$(5x – 1).e^5x + c.

 

h. $\mathop \smallint \nolimits \left( {2{\rm{x}} - 1} \right)$.e^2x.dx

Solution:

Or, $\mathop \smallint \nolimits \left( {2{\rm{x}} - 1} \right)$.e^2x.dx= (2x – 1).$\mathop \smallint \nolimits {{\rm{e}}^{2{\rm{x}}}}.{\rm{dx}} - \mathop \smallint \nolimits \{ \frac{{\rm{d}}}{{{\rm{dx}}}}\left( {2{\rm{x}} - 1} \right)\mathop \smallint \nolimits {{\rm{e}}^{2{\rm{x}}}}.{\rm{dx}}\} $.dx

= (2x – 1). $\frac{{{{\rm{e}}^{2{\rm{x}}}}}}{2}$ – $\mathop \smallint \nolimits 2.\frac{{{{\rm{e}}^{2{\rm{x}}}}}}{2}$.dx = $\frac{1}{2}$(2x – 1).e^2x – $\frac{1}{2}$e^2x + c.

= $\frac{1}{2}$(2x – 1 – 1).e^2x + c = $\frac{1}{2}$(2x – 2)e^2x + c

= (x – 1).e^2x + c.

 

(i) $\mathop \smallint \nolimits {\rm{x}}.{\rm{sinx}}$dx
Solution:

Or, $\mathop \smallint \nolimits {\rm{x}}.{\rm{sinx}}$dx

= x$\mathop \smallint \nolimits {\rm{sinx}}$.dx – $\mathop \smallint \nolimits \{ \frac{{\rm{d}}}{{{\rm{dx}}}}\left( {\rm{x}} \right)\mathop \smallint \nolimits {\rm{sinx}}.{\rm{dx}}\} {\rm{\: }}$.dx

= x(–cosx) – $\mathop \smallint \nolimits 1.\left( { - {\rm{cosx}}} \right).$dx = – xcosx + sinx + c.

 

(j) $\mathop \smallint \nolimits {\rm{xse}}{{\rm{c}}^2}{\rm{x}}$

Solution:

Or, $\mathop \smallint \nolimits {\rm{xse}}{{\rm{c}}^2}{\rm{x}}$= x$\mathop \smallint \nolimits {\sec ^2}{\rm{x}}.{\rm{dx}}$ – $\mathop \smallint \nolimits \{ \frac{{\rm{d}}}{{{\rm{dx}}}}\left( {\rm{x}} \right)\mathop \smallint \nolimits {\sec ^2}{\rm{x}}.{\rm{dx}}\} $.dx

= xtanx – $\mathop \smallint \nolimits 1.{\rm{tanx}}$.dx

= xtanx – $\mathop \smallint \nolimits \frac{{{\rm{sinx}}}}{{{\rm{cosx}}}}$.dx …..(i)

Put cosx = y ⇒ –sinx.dx = dy

So, sinx.dx = –dy

Or, $\mathop \smallint \nolimits \frac{{{\rm{sinx}}}}{{{\rm{cosx}}}}$.dx = $\mathop \smallint \nolimits  - \frac{{{\rm{dy}}}}{{\rm{y}}}$ = –logy – c = – log(cosx) – c

From (i) $\mathop \smallint \nolimits {\rm{x}}.{\sec ^2}{\rm{x}}$.dx = xtanx – (– log(cosx)) – c)

= xtanx + log(cosx) + c.

 

k. $\mathop \smallint \nolimits {\rm{x}}.{\rm{secx}}.{\rm{tanx}}$.dx

Solution:

Or, $\mathop \smallint \nolimits {\rm{x}}.{\rm{secx}}.{\rm{tanx}}$.dx= x$\mathop \smallint \nolimits {\rm{secx}}.{\rm{tanx}}.$dx – $\mathop \smallint \nolimits \{ \frac{{\rm{d}}}{{{\rm{dx}}}}\left( {\rm{x}} \right)\mathop \smallint \nolimits {\rm{secx}}.{\rm{tanx}}\} $.dx

= x.secx – $\mathop \smallint \nolimits 1.{\rm{secx}}$.dx = xsecx –$\mathop \smallint \nolimits {\rm{secx}}$.dx

 = x secx. – log(secx + tanx) + c.

 

(l) $\mathop \smallint \nolimits {\rm{x}}.{\rm{cosnx}}$.dx

Solution:

Or, $\mathop \smallint \nolimits {\rm{x}}.{\rm{cosnx}}$.dx= x$\mathop \smallint \nolimits {\rm{cosnx}}.$dx – $\mathop \smallint \nolimits \{ \frac{{\rm{d}}}{{{\rm{dx}}}}\left( {\rm{x}} \right)\mathop \smallint \nolimits {\rm{cosnx}}.{\rm{dx}}\} $.dx

= x${\rm{\: }}\mathop \smallint \nolimits \frac{{{\rm{sin}}.{\rm{nx}}}}{{\rm{n}}}$ – $\mathop \smallint \nolimits 1.\frac{{{\rm{sinnx}}}}{{\rm{n}}}$.dx = $\frac{1}{{\rm{n}}}$x.sinnx – $\frac{1}{{\rm{n}}}$.$\frac{{1\left( { - {\rm{cosnx}}} \right)}}{{\rm{n}}}$ + c.

= $\frac{1}{{\rm{n}}}$.x.sinnx + $\frac{1}{{{{\rm{n}}^2}}}$cosnx + c

 

m. $\mathop \smallint \nolimits {\rm{x}}.{\rm{sinax}}$.dx

Solution:

or, $\mathop \smallint \nolimits {\rm{x}}.{\rm{sinax}}$.dx= x$\mathop \smallint \nolimits {\rm{sinax}}$.dx –${\rm{\: }}\mathop \smallint \nolimits \{ \frac{{\rm{d}}}{{{\rm{dx}}}}\left( {\rm{x}} \right)\mathop \smallint \nolimits {\rm{sinax}}.{\rm{dx}}\} $.dx

= x.$\frac{{ - {\rm{cosax}}}}{{\rm{a}}}$ – $\mathop \smallint \nolimits 1.\frac{{ - {\rm{cosax}}}}{{\rm{a}}}.{\rm{dx}}$ = $ - \frac{1}{{\rm{a}}}$.x.cosax + $\frac{1}{{\rm{a}}}$.$\frac{{{\rm{sinax}}}}{{\rm{a}}}$ + c

= $\frac{1}{{{{\rm{a}}^2}}}$sinax – $\frac{1}{{\rm{a}}}$x.cosax + c.

 

n. $\mathop \smallint \nolimits {\rm{x}}.{\rm{cos}}\left( {{\rm{ax}} - {\rm{b}}} \right)$.dx

Solution:

or, $\mathop \smallint \nolimits {\rm{x}}.{\rm{cos}}\left( {{\rm{ax}} - {\rm{b}}} \right)$.dx= x$\mathop \smallint \nolimits {\rm{cos}}\left( {{\rm{ax}} - {\rm{b}}} \right)$.dx –${\rm{\: }}\mathop \smallint \nolimits \{ \frac{{\rm{d}}}{{{\rm{dx}}}}\left( {\rm{x}} \right)\mathop \smallint \nolimits {\rm{cos}}\left( {{\rm{ax}} - {\rm{b}}} \right).{\rm{dx}}\} $.dx

= x.$\frac{{{\rm{sin}}\left( {{\rm{ax}} - {\rm{b}}} \right)}}{{\rm{a}}}$ – $\mathop \smallint \nolimits 1.\frac{{\sin \left( {{\rm{ax}} - {\rm{b}}} \right)}}{{\rm{a}}}.{\rm{dx}}$

= $\frac{1}{{\rm{a}}}$.x.sin(ax – b) – $\frac{1}{{\rm{a}}}\frac{{ - \cos \left( {{\rm{ax}} - {\rm{b}}} \right)}}{{\rm{a}}}$ + c

= $\frac{1}{{{{\rm{a}}^2}}}$x.sin(ax – b) + $\frac{1}{{{{\rm{a}}^2}}}$ cos(ax – b) + c.

 

o. $\mathop \smallint \nolimits {{\rm{x}}^2}.{\rm{sinx}}$.dx

Solution:

Or, $\mathop \smallint \nolimits {{\rm{x}}^2}.{\rm{sinx}}$.dx= x².$\mathop \smallint \nolimits {\rm{sinx}}.$dx – $\mathop \smallint \nolimits \{ \frac{{\rm{d}}}{{{\rm{dx}}}}\left( {{{\rm{x}}^2}} \right)\mathop \smallint \nolimits {\rm{sinx}}.{\rm{dx}}\} $.dx

= x².(–cosx) –$\mathop \smallint \nolimits 2{\rm{x}}.\left( { - {\rm{cosx}}} \right)$.dx = –x²cosx + 2$\mathop \smallint \nolimits {\rm{x}}.{\rm{cosx}}$.dx

= –x²cosx + 2 $[{\rm{x}}\mathop \smallint \nolimits {\rm{cosx}}.{\rm{dx}} - \mathop \smallint \nolimits \{ \frac{{\rm{d}}}{{{\rm{dx}}}}\left( {\rm{x}} \right)\mathop \smallint \nolimits {\rm{cosx}}.{\rm{dx}}\} {\rm{dx}}]{\rm{\: }}$

= –x².cosx + 2[x.sinx – $\mathop \smallint \nolimits 1.{\rm{sinxdx}}$]

= –x²cosx + 2.x.sinx – 2(–cosx) + c

= –x².cosx + 2xsinx + 2cosx + c.

 

p. $\mathop \smallint \nolimits {{\rm{x}}^2}.{\rm{cosnx}}$.dx

Solution:

Or, $\mathop \smallint \nolimits {{\rm{x}}^2}.{\rm{cosnx}}$.dx= x².$\mathop \smallint \nolimits {\rm{cosnx}}.$dx – $\mathop \smallint \nolimits \{ \frac{{\rm{d}}}{{{\rm{dx}}}}\left( {{{\rm{x}}^2}} \right)\mathop \smallint \nolimits {\rm{cosnx}}.{\rm{dx}}\} $.dx

= x². $\frac{{{\rm{sinnx}}}}{{\rm{n}}}$ –$\mathop \smallint \nolimits 2{\rm{x}}.\frac{{{\rm{sinnx}}}}{{\rm{n}}}$.dx

= $\frac{1}{2}$x²sinnx – $\frac{2}{{\rm{n}}}$$\left[ {{\rm{x}}\mathop \smallint \nolimits {\rm{sinnx}}.{\rm{dx}} - \left\{ {\mathop \smallint \nolimits \frac{{\rm{d}}}{{{\rm{dx}}}}\left( {\rm{x}} \right)\mathop \smallint \nolimits {\rm{sinnx}}.{\rm{dx}}} \right\}} \right]$

= $\frac{1}{{\rm{n}}}$ x².sinnx – $\frac{2}{{\rm{n}}}$$\left[ {{\rm{x}}.\frac{{ - {\rm{cosnx}}}}{{\rm{n}}} - \mathop \smallint \nolimits 1.\frac{{ - {\rm{cosnx}}}}{{\rm{n}}}.{\rm{dx\: }}} \right]$

= $\frac{1}{{\rm{n}}}$.x².sinnx + $\frac{2}{{{{\rm{n}}^2}}}$x.cosnx – $\frac{2}{{{{\rm{n}}^2}}}$.$\frac{{{\rm{sinnx}}}}{{\rm{n}}}$ + c.

= $\frac{1}{{\rm{n}}}$x².sinnx + $\frac{2}{{{{\rm{n}}^2}}}$x.cosnx – $\frac{2}{{{{\rm{n}}^3}}}$.sinnx + c.

 

q. $\mathop \smallint \nolimits {{\rm{x}}^2}.{\rm{sin}}\left( {3{\rm{x}} - 2} \right)$.dx

Solution:

Or, $\mathop \smallint \nolimits {{\rm{x}}^2}.{\rm{sin}}\left( {3{\rm{x}} - 2} \right)$.dx= x².$\mathop \smallint \nolimits {\rm{sin}}\left( {3{\rm{x}} - 2} \right).$dx – $\mathop \smallint \nolimits \{ \frac{{\rm{d}}}{{{\rm{dx}}}}\left( {{{\rm{x}}^2}} \right)\mathop \smallint \nolimits {\rm{sin}}\left( {3{\rm{x}} - 2} \right).{\rm{dx}}\} $.dx

= x². $\frac{{\left( { - \cos \left( {3{\rm{x}} - 2} \right)} \right)}}{3}$ – $\mathop \smallint \nolimits 2{\rm{x}}.\frac{{\left( { - \cos \left( {3{\rm{x}} - 2} \right)} \right)}}{3}$.dx

= $ - \frac{1}{3}$x².cos(3x – 2) + $\frac{2}{3}$$\left[ {{\rm{x}}.\frac{{\sin \left( {3{\rm{x}} - 2} \right)}}{3} - \mathop \smallint \nolimits 1.\frac{{\sin \left( {3{\rm{x}} - 2} \right)}}{3}.{\rm{dx}}} \right]$

= $ - \frac{1}{3}$x².cos(3x – 2) + $\frac{2}{9}$.x.sin(3x – 2) – $\frac{2}{9}\frac{{ - \cos \left( {3{\rm{x}} - 2} \right)}}{3}$ + c

= $ - \frac{1}{3}$x² cos(3x – 2) + $\frac{2}{9}$.x.sin(3x – 2) + $\frac{2}{{27}}$cos(3x – 2) + c.

 

r. $\mathop \smallint \nolimits {\rm{x}}.{\sin ^2}{\rm{x}}$.dx

Solution:

Or, $\mathop \smallint \nolimits {\rm{x}}.{\sin ^2}{\rm{x}}$.dx= $\frac{1}{2}$.$\mathop \smallint \nolimits {\rm{x}}.\left( {2{{\sin }^2}{\rm{x}}} \right){\rm{dx}}.$ = $\frac{1}{2}\mathop \smallint \nolimits {\rm{x}}\left( {1 - {\rm{cos}}2{\rm{x}}} \right)$.dx

= $\frac{1}{2}\mathop \smallint \nolimits \left( {{\rm{x}} - {\rm{xcos}}2{\rm{x}}} \right)$.dx = $\frac{1}{2}\mathop \smallint \nolimits {\rm{x}}$.dx – $\frac{1}{2}\mathop \smallint \nolimits {\rm{x}}.{\rm{cos}}2{\rm{x}}$.dx

= $\frac{1}{2}.\frac{{{{\rm{x}}^2}}}{2}$ – $\frac{1}{2}$$[{\rm{x}}\mathop \smallint \nolimits {\rm{cos}}2{\rm{x}}.{\rm{dx}} - \mathop \smallint \nolimits \{ \frac{{\rm{d}}}{{{\rm{dx}}}}\left( {\rm{x}} \right)\mathop \smallint \nolimits {\rm{cos}}2{\rm{x}}.{\rm{dx}}\} {\rm{\: }}.{\rm{dx}}]$

= $\frac{1}{4}$x² – $\frac{1}{2}$$\left[ {\frac{{{\rm{xsin}}2{\rm{x}}}}{2} - \mathop \smallint \nolimits 1.\frac{{{\rm{sin}}2{\rm{x}}}}{2}{\rm{dx}}} \right]$

= $\frac{1}{4}$x² – $\frac{1}{4}$xsin2x + $\frac{1}{4}$.$\frac{{ - {\rm{cos}}2{\rm{x}}}}{2}$ + c

= $\frac{1}{4}$x² – $\frac{1}{4}$x.sin2x – $\frac{1}{8}$cos2x + c.

 

s. $\mathop \smallint \nolimits {\rm{x}}.{\cos ^2}2{\rm{x}}$.dx

Solution:

Or, $\mathop \smallint \nolimits {\rm{x}}.{\cos ^2}2{\rm{x}}$.dx= $\frac{1}{2}\mathop \smallint \nolimits {\rm{x}}.(2{\cos ^2}2{\rm{x}}).{\rm{dx}}$ = $\frac{1}{2}\mathop \smallint \nolimits {\rm{x}}\left( {1 + {\rm{cos}}4{\rm{x}}} \right).$dx

= $\frac{1}{2}$$\mathop \smallint \nolimits ({\rm{x}} + {\rm{xcos}}4{\rm{x}}).{\rm{dx}}$ = $\frac{1}{2}\mathop \smallint \nolimits {\rm{x}}.{\rm{dx}} + \frac{1}{2}\mathop \smallint \nolimits {\rm{x}}.{\rm{cos}}4{\rm{x}}$.dx

= $\frac{1}{2}$.$\frac{{{{\rm{x}}^2}}}{2}$ + $\frac{1}{2}$$[{\rm{x}}\mathop \smallint \nolimits {\rm{cos}}4{\rm{x}}.{\rm{dx}} - \mathop \smallint \nolimits \{ \frac{{\rm{d}}}{{{\rm{dx}}}}\left( {\rm{x}} \right)\mathop \smallint \nolimits {\rm{cos}}4{\rm{x}}.{\rm{dx}}\} {\rm{\: }}.{\rm{dx}}]{\rm{\: }}$

= $\frac{1}{4}$x² + $\frac{1}{2}$$[{\rm{x}}.\frac{{{\rm{sin}}4{\rm{x}}}}{4} - \mathop \smallint \nolimits 1.\frac{{{\rm{sin}}4{\rm{x}}}}{4}$.dx$]$

= $\frac{1}{4}$x² + $\frac{1}{8}$x.sin4x – $\frac{1}{8}$. $\frac{{ - {\rm{cos}}4{\rm{x}}}}{4}$ + c

= $\frac{1}{4}$x² + $\frac{1}{8}$x.sin4x + $\frac{1}{{32}}$.cos4x + c.

 

t. $\mathop \smallint \nolimits {\rm{x}}.{\sin ^2}{\rm{nx}}$.dx

Solution:

Or, $\mathop \smallint \nolimits {\rm{x}}.{\sin ^2}{\rm{nx}}$.dx= $\frac{1}{2}\mathop \smallint \nolimits {\rm{x}}(2{\sin ^2}{\rm{nx}}).{\rm{dx}}$ = $\frac{1}{2}\mathop \smallint \nolimits {\rm{x}}\left( {1 - {\rm{cos}}2{\rm{nx}}} \right).{\rm{dx}}$

= $\frac{1}{2}\mathop \smallint \nolimits \left( {{\rm{x}} - {\rm{xcos}}2{\rm{nx}}} \right).{\rm{dx}}$ = $\frac{1}{2}\mathop \smallint \nolimits {\rm{x}}.{\rm{dx}}$ – $\frac{1}{2}\mathop \smallint \nolimits {\rm{x}}.{\rm{cos}}2{\rm{nx}}$.dx

= $\frac{1}{2}$.$\frac{{{{\rm{x}}^2}}}{2}$ – $\frac{1}{2}[{\rm{x}}\mathop \smallint \nolimits {\rm{cos}}2{\rm{nx}}.{\rm{dx}} - \mathop \smallint \nolimits \{ \frac{{\rm{d}}}{{{\rm{dx}}}}\left( {\rm{x}} \right)\mathop \smallint \nolimits {\rm{cos}}2{\rm{nx}}.{\rm{dx}}\} {\rm{\: dx}}]$

= $\frac{1}{4}$x² – $\frac{1}{2}$$\left[ {{\rm{x}}.\frac{{{\rm{sin}}2{\rm{nx}}}}{{2{\rm{n}}}} - \mathop \smallint \nolimits 1.\frac{{{\rm{sin}}2{\rm{nx}}}}{{2{\rm{x}}}}.{\rm{dx}}} \right]$

= $\frac{1}{4}$x² – $\frac{1}{{4{\rm{n}}}}$x.sin2nx + $\frac{1}{{4{\rm{n}}}}$$\mathop \smallint \nolimits {\rm{sin}}2{\rm{nx}}$.dx

= $\frac{1}{4}$x² – $\frac{1}{{4{\rm{n}}}}$x.sin2n + $\frac{1}{{4{\rm{n}}}}$. $\frac{{ - {\rm{cos}}2{\rm{nx}}}}{{2{\rm{n}}}}$ + c

= $\frac{1}{4}$x² – $\frac{1}{{4{\rm{n}}}}$x.sin2nx – $\frac{1}{{8{{\rm{n}}^2}}}$cos2nx + c.

 

u. $\mathop \smallint \nolimits {\rm{x}}.{\tan ^2}{\rm{nx}}$.dx

Solution:

Or, $\mathop \smallint \nolimits {\rm{x}}.{\tan ^2}{\rm{nx}}$.dx= $\mathop \smallint \nolimits {\rm{x}}({\sec ^2}{\rm{nx}} - 1).{\rm{dx}}$ = $\mathop \smallint \nolimits \left( {{\rm{xse}}{{\rm{c}}^2}{\rm{nx}} - {\rm{x}}} \right).{\rm{dx}}$

= $\mathop \smallint \nolimits {\rm{x}}.{\sec ^2}{\rm{nx}}$.dx – $\mathop \smallint \nolimits {\rm{x}}.{\rm{dx}}$

= ${\rm{x}}\mathop \smallint \nolimits {\sec ^2}{\rm{nx}}$.dx – $\mathop \smallint \nolimits \left\{ {\frac{{\rm{d}}}{{{\rm{dx}}}}\left( {\rm{x}} \right)\mathop \smallint \nolimits {{\sec }^2}{\rm{nx}} - {\rm{dx}}} \right\}.{\rm{dx}} - \frac{{{{\rm{x}}^2}}}{2}$

= x. $\frac{{{\rm{tannx}}}}{{\rm{n}}}$ – $\mathop \smallint \nolimits 1.\frac{{{\rm{tannx}}}}{{\rm{n}}}.{\rm{dx}}$ – $\frac{{{{\rm{x}}^2}}}{2}$.

= $\frac{1}{{\rm{n}}}$.x.tan.nx – $\frac{1}{{\rm{n}}}$$\mathop \smallint \nolimits {\rm{tannx}}.{\rm{dx}}$ – $\frac{{{{\rm{x}}^2}}}{2}$.

= $\frac{1}{{\rm{n}}}$.xtannx – $\frac{1}{{\rm{n}}}\mathop \smallint \nolimits \frac{{{\rm{secnx}}.{\rm{tannx}}}}{{{\rm{secnx}}}}$.dx – $\frac{{{{\rm{x}}^2}}}{2}$

= $\frac{1}{{\rm{n}}}$.x.tanx – $\frac{1}{{\rm{n}}}$.$\frac{1}{{\rm{n}}}$.log secnx – $\frac{{{{\rm{x}}^2}}}{2}$ + c

= $\frac{1}{{\rm{n}}}$.x.tannx – $\frac{1}{{{{\rm{n}}^2}}}{\rm{\: }}$log(cosnx)^–1 – $\frac{1}{2}$x² + c

= $\frac{1}{{\rm{n}}}$x.tan.nx + $\frac{1}{{{{\rm{n}}^2}}}$log.cosnx – $\frac{1}{2}$x² + c.

 

(v) $\mathop \smallint \nolimits {\sec ^3}{\rm{x}}$.dx

Solution:

Or, $\mathop \smallint \nolimits {\sec ^3}{\rm{x}}$.dx= $\mathop \smallint \nolimits {\rm{secx}}.{\sec ^2}{\rm{x}}$.dx

= secx$\mathop \smallint \nolimits {\sec ^2}{\rm{x}}$.dx – $\mathop \smallint \nolimits \{ \frac{{\rm{d}}}{{{\rm{dx}}}}\left( {{\rm{secx}}} \right)\mathop \smallint \nolimits {\sec ^2}{\rm{x}}.{\rm{dx}}\} $.dx

= secx.tanx – $\mathop \smallint \nolimits {\rm{secx}}.{\rm{tanx}}.{\rm{tanx}}.$dx

= secx.tanx – $\mathop \smallint \nolimits {\rm{secx}}\left( {{{\sec }^2}{\rm{x}} - 1} \right)$.dx

= secx.tanx – $\mathop \smallint \nolimits ({\sec ^3}{\rm{x}} - \sec {\rm{x}})$.dx

Or, $\mathop \smallint \nolimits {\sec ^3}{\rm{x}}$.dx = secx.tanx – $\mathop \smallint \nolimits {\sec ^3}{\rm{x}}$.dx + $\mathop \smallint \nolimits {\rm{secx}}$.dx

Or, 2 $\mathop \smallint \nolimits {\sec ^3}{\rm{x}}$.dx = secx.tanx + $\mathop \smallint \nolimits \frac{{{\rm{secx}}\left( {{\rm{secx}} - {\rm{tanx}}} \right)}}{{{\rm{secx}} + {\rm{tanx}}}}$.dx

Or, 2$\mathop \smallint \nolimits {\sec ^3}{\rm{x}}$.dx = secx.tanx + log(secx + tanx) + 2c.

Or, $\mathop \smallint \nolimits {\sec ^3}{\rm{x}}$.dx = $\frac{1}{2}$.secx.tanx + $\frac{1}{2}$.log(secx + tanx) + c.

 

(w) $\mathop \smallint \nolimits {\rm{cose}}{{\rm{c}}^3}{\rm{x}}$.dx

Solution:

Or, $\mathop \smallint \nolimits {\rm{cose}}{{\rm{c}}^3}{\rm{x}}$.dx= $\mathop \smallint \nolimits {\rm{cosecx}}.{\rm{cose}}{{\rm{c}}^2}{\rm{x}}$.dx

= cosec.x$\mathop \smallint \nolimits {\rm{cose}}{{\rm{c}}^2}{\rm{x}}$.dx – $\mathop \smallint \nolimits \{ \frac{{\rm{d}}}{{{\rm{dx}}}}\left( {{\rm{cosecx}}} \right)\mathop \smallint \nolimits {\rm{cose}}{{\rm{c}}^2}{\rm{x}}.{\rm{dx}}\} $.dx

= cosec.x.(–cotx) – $\mathop \smallint \nolimits  - {\rm{cosecx}}.{\rm{cotx}}.\left( { - {\rm{cotx}}} \right).$dx

= $ - {\rm{cosecx}}.{\rm{cotx}}$ – $\mathop \smallint \nolimits {\rm{cose}}{{\rm{c}}^2}{\rm{x}}.{\rm{cos}}{{\rm{t}}^2}{\rm{x}}$.dx

= $ - {\rm{cosecx}}.{\rm{cotx}}$ – $\mathop \smallint \nolimits {\rm{cosecx}}\left( {{\rm{cose}}{{\rm{c}}^2}{\rm{x}} - 1} \right)$.dx

= $ - {\rm{cosecx}}.{\rm{cotx}}$ – $\mathop \smallint \nolimits \left( {{\rm{cose}}{{\rm{c}}^3}{\rm{x}} - {\rm{cosecx}}} \right)$.dx

= $\mathop \smallint \nolimits {\rm{cose}}{{\rm{c}}^3}{\rm{x}}.{\rm{dx}}$ = $ - {\rm{cosecx}}.{\rm{cotx}}$$--$$\mathop \smallint \nolimits {\rm{cose}}{{\rm{c}}^3}{\rm{x}}.$dx + $\mathop \smallint \nolimits {\rm{cosec}}$.dx

Or, 2 $\mathop \smallint \nolimits {\rm{cose}}{{\rm{c}}^3}{\rm{x}}$.dx = $ - {\rm{cosecx}}.{\rm{cotx}}$ + $\mathop \smallint \nolimits \frac{{{\rm{cosecx}}\left( {{\rm{cosecx}} - {\rm{cotx}}} \right)}}{{{\rm{cosecx}} - {\rm{cotx}}}}$.dx

= $ - {\rm{cosecx}}$.cotx + log(cosecx – cotx) + 2c.

So, $\mathop \smallint \nolimits {\rm{cose}}{{\rm{c}}^3}{\rm{x}}$.dx = $ - \frac{1}{2}$.cosecx.cotx + $\frac{1}{2}$log(cosecx – cotx) + c.

 

(x) $\mathop \smallint \nolimits {{\rm{e}}^{\rm{x}}}.{\rm{sinx}}$.dx

Solution:

Or, $\mathop \smallint \nolimits {{\rm{e}}^{\rm{x}}}.{\rm{sinx}}$.dx= ${{\rm{e}}^{{\rm{ax}}}}\mathop \smallint \nolimits {\rm{sinbx}}. - \mathop \smallint \nolimits \{ \frac{{\rm{d}}}{{{\rm{dx}}}}\left( {{{\rm{e}}^{{\rm{ax}}}}} \right)\mathop \smallint \nolimits {\rm{sinbx}}\} $.dx

= ${{\rm{e}}^{{\rm{ax}}}}$$\frac{{ - {\rm{cosbx}}}}{{\rm{b}}}$ – $\mathop \smallint \nolimits {\rm{a}}.{{\rm{e}}^{{\rm{ax}}}}.\frac{{ - {\rm{cosbx}}}}{{\rm{b}}}$. .dx

= $ - \frac{{{{\rm{e}}^{{\rm{ax}}}}.{\rm{cosbx}}}}{{\rm{b}}}$ + $\frac{{\rm{a}}}{{\rm{b}}}$$[{{\rm{e}}^{{\rm{ax}}}}\mathop \smallint \nolimits {\rm{cosbx}}.{\rm{dx}} - \mathop \smallint \nolimits \{ \frac{{\rm{d}}}{{{\rm{dx}}}}\left( {{{\rm{e}}^{{\rm{ax}}}}} \right)\mathop \smallint \nolimits {\rm{cosbx}}.{\rm{dx}}\} .{\rm{\: dx}}]$.

= $ - \frac{{{{\rm{e}}^{{\rm{ax}}}}.{\rm{cosbs}}}}{{\rm{b}}}$ + $\frac{{\rm{a}}}{{\rm{b}}}$$\left[ {{{\rm{e}}^{{\rm{ax}}}}\frac{{{\rm{sinbx}}}}{{\rm{b}}} - \mathop \smallint \nolimits {\rm{a}}{{\rm{e}}^{{\rm{ax}}}}.\frac{{{\rm{sinbx}}}}{{\rm{b}}}.{\rm{dx}}} \right]$ + c₁.

Or, $\mathop \smallint \nolimits {{\rm{e}}^{{\rm{ax}}}}{\rm{sinbx}}$.dx = $\frac{{{{\rm{e}}^{{\rm{ax}}}}.{\rm{cosbx}}}}{{\rm{b}}}$ + $\frac{{\rm{a}}}{{{{\rm{b}}^2}}}$e^ax.sinbx – $\frac{{{{\rm{a}}^2}}}{{{{\rm{b}}^2}}}$$\mathop \smallint \nolimits {{\rm{e}}^{{\rm{ax}}}}$.sinbx.dx + c₁

Or, $\mathop \smallint \nolimits {{\rm{e}}^{{\rm{ax}}}}$.sinbx.dx + $\frac{{{{\rm{a}}^2}}}{{{{\rm{b}}^2}}}$$\mathop \smallint \nolimits {{\rm{e}}^{{\rm{ax}}}}$.sinbx.dx

= $\frac{{ - {{\rm{e}}^{{\rm{ax}}}}.{\rm{bcosbx}} + {\rm{a}}{{\rm{e}}^{{\rm{ax}}}}.{\rm{sinbx}}}}{{{{\rm{b}}^2}}}$ + c₁

Or, $\left( {1 + \frac{{{{\rm{a}}^2}}}{{{{\rm{b}}^2}}}} \right)\mathop \smallint \nolimits {{\rm{e}}^{{\rm{ax}}}}.{\rm{sinbx}}.{\rm{dx}}$ = $\frac{{\left( {{\rm{asinbx}} - {\rm{bcosbx}}} \right).{{\rm{e}}^{{\rm{ax}}}}}}{{{{\rm{b}}^2}}}$ + c₁

Or, $\frac{{{{\rm{a}}^2} + {{\rm{b}}^2}}}{{{{\rm{b}}^2}}}$$\mathop \smallint \nolimits {{\rm{e}}^{{\rm{ax}}}}.{\rm{sinbx}}.{\rm{dx}}$ = $\frac{{\left( {{\rm{asinbx}} - {\rm{b}}.{\rm{cosbx}}} \right).{{\rm{e}}^{{\rm{ax}}}}}}{{{{\rm{b}}^2}}}$ + c₁.

Or, $\mathop \smallint \nolimits {{\rm{e}}^{{\rm{ax}}}}$.sinbx.dx = $\frac{{\left( {{\rm{a}}.{\rm{sinbx}} - {\rm{b}}.{\rm{cosbx}}} \right){{\rm{e}}^{{\rm{ax}}}}}}{{{{\rm{a}}^2} + {{\rm{b}}^2}}}$ + $\frac{{{{\rm{b}}^2}}}{{{{\rm{a}}^2} + {{\rm{b}}^2}}}$.c₁

= $\frac{{\left( {{\rm{asinbx}} - {\rm{bcosbx}}} \right).{{\rm{e}}^{{\rm{ax}}}}}}{{{{\rm{a}}^2} + {{\rm{b}}^2}}}$ + c, where c = $\frac{{{{\rm{b}}^2}}}{{{{\rm{a}}^2} + {{\rm{b}}^2}}}$.c₁.

$(y)\; \int e^{ax}\sin bxdx$

$Solution:$

$I=\int e^{ax}\sin bxdxIntegrate\; by\; parts\; -=e^{ax}\int \sin bxdx-\int \frac{d{e}^{ax}}{dx}\left(\int \sin bxdx\right)dx=e^{ax}\frac{-cosbx}{b}-\int a{e}^{ax}\frac{-\cos bx}{b}dx=-e^{ax}\frac{\cos bx}{b}+\frac{a}{b}\int e^{ax}\cos bxdx=\frac{e^{ax}}{b}\cos bx+\frac{a}{b}\left[e^{ax}\int \cos bxdx-\int \frac{d{e}^{ax}}{dx}\left(\int \cos bxdx\right)dx\right]=\frac{e^{ax}}{b}\cos bx+\frac{a}{b}\left[e^{ax}\frac{\sin bx}{b}-\int a{e}^{ax}\sin bxdx\right]=\frac{-e^{ax}}{b}\cos bx+\frac{a{e}^{ax}}{b^2}\sin bx-\frac{a^2}{b^2}\int e^{ax}\sin bxdxI=e^{ax}\frac{-\cos bx}{b}+\frac{a{e}^{ax}}{b^2}\sin bx-I\left(\frac{a^2}{b^2}\right)I+\frac{a^2}{b^2}I=e^{ax}\frac{\left(a\sin bx-b\cos bs\right)}{b^2}I\frac{a^2+b^2}{b^2}=e^{ax}\frac{\left(a\sin bx-b\cos bx\right)}{b^2}I=\frac{e^{ax}}{a^2+b^2}\left(a\sin bx-b\cos bx\right)$

z.$\mathop \smallint \nolimits {{\rm{e}}^{\rm{x}}}.{\rm{cosbx}}$.dx

Solution:

Or, $\mathop \smallint \nolimits {{\rm{e}}^{\rm{x}}}.{\rm{cosbx}}$.dx= ${{\rm{e}}^{{\rm{ax}}}}\mathop \smallint \nolimits {\rm{cos}}.{\rm{bx}}.{\rm{dx}} - \mathop \smallint \nolimits \{ \frac{{\rm{d}}}{{{\rm{dx}}}}\left( {{{\rm{e}}^{{\rm{ax}}}}} \right)\mathop \smallint \nolimits {\rm{cosbx}}\} $.dx

= ${{\rm{e}}^{{\rm{ax}}}}$$\frac{{{\rm{sinbx}}}}{{\rm{b}}}$ – $\mathop \smallint \nolimits {\rm{a}}.{{\rm{e}}^{{\rm{ax}}}}.\frac{{{\rm{sinbx}}}}{{\rm{b}}}$.dx

= $\frac{1}{{\rm{b}}}$e^ax.sinbx – $\frac{{\rm{a}}}{{\rm{b}}}$$\mathop \smallint \nolimits {{\rm{e}}^{{\rm{ax}}}}.{\rm{sinbx}}$.dx

= $\frac{1}{{\rm{b}}}$.e^ax.sinbx – $\frac{{\rm{a}}}{{\rm{b}}}$$\left[ {{{\rm{e}}^{{\rm{ax}}}}\mathop \smallint \nolimits {\rm{sinbx}}.{\rm{dx}} - \mathop \smallint \nolimits \{ \frac{{\rm{d}}}{{{\rm{dx}}}}\left( {{{\rm{e}}^{{\rm{ax}}}}} \right)\mathop \smallint \nolimits {\rm{sinbx}}.{\rm{dx}}\} .{\rm{dx}}} \right]$

= $\frac{1}{{\rm{b}}}$.e^ax.sinb – $\frac{{\rm{a}}}{{\rm{b}}}$$\left[ {{{\rm{e}}^{{\rm{ax}}}}.\frac{{ - {\rm{cosbx}}}}{{\rm{b}}} - \mathop \smallint \nolimits {\rm{a}}{{\rm{e}}^{{\rm{ax}}}}.\left( {\frac{{ - {\rm{cosbx}}}}{{\rm{b}}}} \right).{\rm{dx}}} \right]$ + c₁

= $\frac{1}{{\rm{b}}}$e^ax.sinbx + $\frac{{\rm{a}}}{{{{\rm{b}}^2}}}$.e^ax.cosbx – $\frac{{{{\rm{a}}^2}}}{{{{\rm{b}}^2}}}\mathop \smallint \nolimits {{\rm{e}}^{{\rm{ax}}}}.{\rm{cosbxdx}}$ + c₁

Or, $\mathop \smallint \nolimits {{\rm{e}}^{{\rm{ax}}}}.{\rm{cosbx}}$.dx + $\frac{{{{\rm{a}}^2}}}{{{{\rm{b}}^2}}}$$\mathop \smallint \nolimits {{\rm{e}}^{{\rm{ax}}}}.{\rm{cosx}}.{\rm{bx}}$.dx= $\frac{{{\rm{b}}.{{\rm{e}}^{{\rm{ax}}}}.{\rm{sinbx}} + {\rm{a}}{{\rm{e}}^{{\rm{ax}}}}.{\rm{cosbx}}}}{{{{\rm{b}}^2}}}$ + c₁.

Or, $\left( {1 + \frac{{{{\rm{a}}^2}}}{{{{\rm{b}}^2}}}} \right)$$\mathop \smallint \nolimits {{\rm{e}}^{{\rm{ax}}}}{\rm{cosbx}}.{\rm{dx}}$ = $\frac{{{{\rm{e}}^{{\rm{ax}}}}\left( {{\rm{bsinbx}} + {\rm{a}}.{\rm{cosbx}}} \right)}}{{{{\rm{b}}^2}}}$ + c₁

Or, $\frac{{{{\rm{a}}^2} + {{\rm{b}}^2}}}{{{{\rm{b}}^2}}}\mathop \smallint \nolimits {{\rm{e}}^{{\rm{ax}}}}{\rm{cosbx}}.{\rm{dx}}$ = $\frac{{{{\rm{e}}^{{\rm{ax}}}}\left( {{\rm{bsinbx}} + {\rm{acosbx}}} \right)}}{{{{\rm{b}}^2}}}$ + c₁

Or, $\mathop \smallint \nolimits {{\rm{e}}^{{\rm{ax}}}}$cosbx.dx = $\frac{{{{\rm{e}}^{{\rm{ax}}}}\left( {{\rm{bsinbx}} + {\rm{acosbx}}} \right)}}{{{{\rm{a}}^2} + {{\rm{b}}^2}}}$ + $\frac{{{{\rm{b}}^2}}}{{{{\rm{a}}^2} + {{\rm{b}}^2}}}$. C₁

So, $\mathop \smallint \nolimits {{\rm{e}}^{{\rm{ax}}}}$cosbx.dx = $\frac{{{{\rm{e}}^{{\rm{ax}}}}\left( {{\rm{bsinbx}} + {\rm{a}}.{\rm{cosbx}}} \right)}}{{{{\rm{a}}^2} + {{\rm{b}}^2}}}$ + c,

Where, c = $\frac{{{{\rm{b}}^2}}}{{{{\rm{a}}^2} + {{\rm{b}}^2}}}$. c_1.