12th NCERT INTEGRALS Exercise 7.6 Questions 24

Contact US     Exercise :-     7.1     7.2    7.3    7.4    7.5    7.6    7.7    7.8    7.9    7.10    7.11    Miscellaneous Q1 to 24     Miscellaneous Q25 to 44     Extra     maths index

Do or do not
There is no try

Integrate the rational functions in Exercise 1 to 21

Question (1)

xsinx

Solution

\[I = \int {x\sin xdx} \] \[I = \int {x\sin xdx} \] \[I = x\int {\sin xdx - \int {\left( {\frac{d}{{dx}}x\int {\sin xdx} } \right)} } dx\] \[I = x\left( { - \cos x} \right) - \int {1\left( { - \cos x} \right)dx} \] \[I = - x\cos x + \int {\cos xdx} \] \[I = - x\cos x + sinx + c\]

Question (2)

xsin3x

Solution

\[I = \int {x\sin 3xdx} \] \[I = x\int {\sin 3xdx - \int {\left( {\frac{d}{{dx}}x\int {\sin 3xdx} } \right)} } dx\] \[I = x\left( {\frac{{ - \cos 3x}}{3}} \right) - \int {1\left( {\frac{{ - \cos 3x}}{3}} \right)dx} \] \[I = - x\frac{{\cos 3x}}{3} + \frac{1}{3}\int {\cos 3xdx} \] \[I = - \frac{x}{3}\cos x + \frac{1}{9}sin3x + c\]

Question (3)

x²e^x

Solution

\[I = \int {{x^2}{e^x}dx} \] \[I = {x^2}\int {{e^x}dx - \int {\left( {\frac{d}{{dx}}{x^2}\int {{e^x}dx} } \right)} } dx\] \[I = {x^2}{e^x} - \int {2x\left( {{e^x}} \right)dx} \] \[I = {x^2}{e^x} - 2\left[ {x\int {{e^x}dx - \int {\frac{d}{{dx}}x\int {{e^x}} dx} } } \right]\] \[I = {x^2}{e^x} - 2x{e^x} + 2{e^x} + c\] \[I = {e^x}\left( {{x^2} - 2x + 2} \right) + c\]

Question (4)

xlogx

Solution

\[I = \int {x\log xdx} \] \[I = \int {\log x \cdot xdx} \] \[I = \log x\int {xdx - \int {\left( {\frac{d}{{dx}}\log x\int {xdx} } \right)} } dx\] \[I = \log x \cdot \frac{{{x^2}}}{2} - \int {\frac{1}{\require{cancel}\cancel{x}} \cdot \frac{{{\cancel{x^2}^x}}}{2}dx} \] \[I = \frac{{{x^2}\log x}}{2} - \frac{1}{2}\int {xdx} \] \[I = \frac{{{x^2}\log x}}{2} - \frac{{{x^2}}}{4} + c\]

Question (5)

xlog2x

Solution

\[I = \int {x\log 2xdx} \] \[I = \int {\log 2x \cdot xdx} \] \[I = \log 2x\int {xdx - \int {\left( {\frac{d}{{dx}}\log 2x\int {xdx} } \right)} } dx\] \[I = \log 2x \cdot \frac{{{x^2}}}{2} - \int {\frac{1}{{\cancel{2x}}} \times \cancel{2} \cdot \frac{{{\cancel{x^2}x}}}{2}dx} \] \[I = \frac{{{x^2}\log 2x}}{2} - \frac{1}{2}\int {xdx} \] \[I = \frac{{{x^2}\log 2x}}{2} - \frac{1}{2}\frac{{{x^2}}}{2} + c\] \[I = \frac{{{x^2}\log 2x}}{2} - \frac{{{x^2}}}{4} + c\]

Question (6)

x²logx dx

Solution

\[I = \int {{x^2}\log xdx} \] \[I = \int {\log x \cdot {x^2}dx} \] \[I = \log x\int {{x^2}dx} - \int {\left( {\frac{d}{{dx}}\log x\int {{x^2}dx} } \right)} dx\] \[I = \log x\frac{{{x^3}}}{3} - \int {\frac{1}{x}} \cdot \frac{{{x^3}}}{3}dx\] \[I = \frac{{{x^3}\log x}}{3} - \frac{1}{3}\int {{x^2}} dx\] \[I = \frac{{{x^3}\log x}}{3} - \frac{1}{3}\frac{{{x^3}}}{3} + c\] \[I = \frac{{{x^3}\log x}}{3} - \frac{{{x^3}}}{9} + c\]

Question (7)

xsin^-1x

Solution

\[I = \int {x{{\sin }^{ - 1}}xdx} \] \[I = \int {{{\sin }^{ - 1}}x \cdot xdx} \] \[I = {\sin ^{ - 1}}x\int {xdx} - \int {\left( {\frac{d}{{dx}}{{\sin }^{ - 1}}x\int {xdx} } \right)} dx\] \[I = \frac{{{x^2{\sin }^{ - 1}}x}}{2} - \int {\left( {\frac{1}{{\sqrt {1 - {x^2}} }} \times \frac{{{x^2}}}{2}} \right)} dx\] \[I = \frac{{{x^2{\sin }^{ - 1}}x}}{2} - \frac{1}{2}\int {\left( {\frac{{{x^2}}}{{\sqrt {1 - {x^2}} }}} \right)} dx\qquad equation(1) \] \[\qquad Let{I_1} = \int {\left( {\frac{{ - {x^2}}}{{\sqrt {1 - {x^2}} }}} \right)} dx\] Add 1 and subtract 1 from Numerator
\[{I_1} = \int {\left( {\frac{{1 - {x^2} - 1}}{{\sqrt {1 - {x^2}} }}} \right)} dx\] \[{I_1} = \int {\left( {\frac{{1 - {x^2}}}{{\sqrt {1 - {x^2}} }}} \right)} dx + \int {\left( {\frac{{ - 1}}{{\sqrt {1 - {x^2}} }}} \right)} dx\] \[{I_1} = \int {\sqrt {1 - {x^2}} dx - {{\sin }^{ - 1}}x + c} \] \[{I_1} = \frac{{{x}}}{2}\sqrt {1 - {x^2}} + \frac{1}{2}{\sin ^{ - 1}}x - {\sin ^{ - 1}}x + c\] \[{I_1} = \frac{{{x}}}{2}\sqrt {1 - {x^2}} - \frac{1}{2}{\sin ^{ - 1}}x + c\] Sbustituting value of I₁ n equation(1)
\[I = \frac{{{x^2{\sin }^{ - 1}}x}}{2} - \frac{1}{2}\int {\left( {\frac{{{x^2}}}{{\sqrt {1 - {x^2}} }}} \right)} dx\qquad equation(1) \] \[I = \frac{{{x^2}{{\sin }^{ - 1}}x}}{2} + \frac{1}{2}\left[ {\frac{x}{2}\sqrt {1 - {x^2}} - \frac{{{{\sin }^{ - 1}}x}}{2} + c} \right]\] \[I = \frac{{{x^2}{{\sin }^{ - 1}}x}}{2} + \frac{{x\sqrt {1 - {x^2}} }}{4} - \frac{{{{\sin }^{ - 1}}x}}{4} + c\] \[I = \frac{{{{\sin }^{ - 1}}x}}{2}\left( {{x^2} - \frac{1}{2}} \right) + \frac{{x\sqrt {1 - {x^2}} }}{4} + c\] \[I = \frac{{\left( {2{x^2} - 1} \right){{\sin }^{ - 1}}x}}{4} + \frac{{x\sqrt {1 - {x^2}} }}{4} + c\]

Question (8)

xtan^-1x

Solution

\[I = \int {x{{\tan }^{ - 1}}} xdx\] \[I = \int {{{\tan }^{ - 1}}x \cdot xdx} \] \[I = {\tan ^{ - 1}}x\int {xdx} - \int {\left( {\frac{d}{{dx}}{{\tan }^{ - 1}}x\int {xdx} } \right)} dx\] \[I = {\tan ^{ - 1}}x \cdot \frac{{{x^2}}}{2} - \int {\frac{1}{{1 + {x^2}}} \times \frac{{{x^2}}}{2}dx} \] \[I = \frac{{{x^2}{{\tan }^{ - 1}}x}}{2} - \frac{1}{2}\int {\frac{{{x^2}}}{{1 + {x^2}}}dx} \] Add and subtract 1 in numerator
\[I = \frac{{{x^2}{{\tan }^{ - 1}}x}}{2} - \frac{1}{2}\int {\frac{{1 + {x^2} - 1}}{{1 + {x^2}}}dx} \] \[I = \frac{{{x^2}{{\tan }^{ - 1}}x}}{2} - \frac{1}{2}\int {\frac{{1 + {x^2}}}{{1 + {x^2}}}dx} + \frac{1}{2}\int {\frac{1}{{1 + {x^2}}}} dx\] \[I = \frac{{{x^2}{{\tan }^{ - 1}}x}}{2} - \frac{1}{2}\int {dx} + \frac{1}{2}{\tan ^{ - 1}}x + c\] \[I = \frac{{{x^2}{{\tan }^{ - 1}}x}}{2} - \frac{x}{2} + \frac{{{{\tan }^{ - 1}}x}}{2} + c\]

Question (9)

xcos^-1

Solution

\[I = \int {x{{\cos }^{ - 1}}xdx} \] \[I = \int {{{\cos }^{ - 1}}x \cdot xdx} \] \[I = {\cos ^{ - 1}}x\int {xdx} - \int {\left( {\frac{d}{{dx}}{{\cos }^{ - 1}}x\int {xdx} } \right)} dx\] \[I = \frac{{{x^2{\cos }^{ - 1}}x}}{2} - \int {\left( {\frac{{ - 1}}{{\sqrt {1 - {x^2}} }} \times \frac{{{x^2}}}{2}} \right)} dx\] \[I = \frac{{{x^2{\cos }^{ - 1}}x}}{2} - \frac{1}{2}\int {\left( {\frac{{ - {x^2}}}{{\sqrt {1 - {x^2}} }}} \right)} dx\] Add 1 and subtract 1 from Numerator
\[I = \frac{{{x^2{\cos }^{ - 1}}x}}{2} - \frac{1}{2}\int {\left( {\frac{{1 - {x^2} - 1}}{{\sqrt {1 - {x^2}} }}} \right)} dx\] \[I = \frac{{{x^2}{{\cos }^{ - 1}}x}}{2} - \frac{1}{2}\int {\left( {\frac{{1 - {x^2} - 1}}{{\sqrt {1 - {x^2}} }}} \right)} dx\] \[I = \frac{{{x^2}{{\cos }^{ - 1}}x}}{2} - \frac{1}{2}\int {\left( {\frac{{1 - {x^2}}}{{\sqrt {1 - {x^2}} }}} \right)} dx + \frac{1}{2}\int {\frac{1}{{\sqrt {1 - {x^2}} }}dx} \] \[I = \frac{{{x^2}{{\cos }^{ - 1}}x}}{2} - \frac{1}{2}\int {\sqrt {1 - {x^2}} dx} + \frac{1}{2}{\sin ^{ - 1}}x + {c_1}\] \[I = \frac{{{x^2}{{\cos }^{ - 1}}x}}{2} - \frac{1}{2}\left[ {\frac{x}{2}\sqrt {1 - {x^2}} + \frac{1}{2}{{\sin }^{ - 1}}x} \right] + \frac{1}{2}{\sin ^{ - 1}}x + {c_1}\] \[I = \frac{{{x^2}{{\cos }^{ - 1}}x}}{2} - \frac{{x\sqrt {1 - {x^2}} }}{4} - \frac{1}{4}{\sin ^{ - 1}}x + \frac{1}{2}{\sin ^{ - 1}}x + {c_1}\] \[I = \frac{{{x^2}{{\cos }^{ - 1}}x}}{2} - \frac{{x\sqrt {1 - {x^2}} }}{4} + \frac{1}{4}{\sin ^{ - 1}}x + {c_1}\] \[I = \frac{{{x^2}{{\cos }^{ - 1}}x}}{2} - \frac{{x\sqrt {1 - {x^2}} }}{4} + \frac{1}{4}\left( {\frac{\pi }{2} - {{\cos }^{ - 1}}x} \right) + {c_1}\] \[I = \frac{{{x^2}{{\cos }^{ - 1}}x}}{2} - \frac{{x\sqrt {1 - {x^2}} }}{4} + \frac{\pi }{8} - \frac{{{{\cos }^{ - 1}}x}}{4} + {c_1}\] \[c = {c_1} + \frac{\pi }{8}\] \[I = \frac{{\left( {2{x^2} - 1} \right){{\cos }^{ - 1}}x}}{4} - \frac{{x\sqrt {1 - {x^2}} }}{4} + c\]

Question (10)

(sin^-1x)²

Solution

\[I = \int {{{\left( {{{\sin }^{ - 1}}x} \right)}^2}dx} \] \[I = \int {{{\left( {{{\sin }^{ - 1}}x} \right)}^2} \cdot 1dx} \] \[I = {\left( {{{\sin }^{ - 1}}x} \right)^2}\int {1dx} - \int {\left( {\frac{d}{{dx}}{{\left( {{{\sin }^{ - 1}}x} \right)}^2}\int {1dx} } \right)} dx\] \[I = {\left( {{{\sin }^{ - 1}}x} \right)^2}x - \int {\left( {2{{\sin }^{ - 1}}x} \right)\frac{1}{{\sqrt {1 - {x^2}} }}x} dx\] \[I = {\left( {{{\sin }^{ - 1}}x} \right)^2}x - 2\int {\frac{{{{\sin }^{ - 1}}x \cdot x}}{{\sqrt {1 - {x^2}} }}} dx\] \[I = {\left( {{{\sin }^{ - 1}}x} \right)^2}x + {I_1}\qquad equation(1)\] \[let\qquad {I_1} = \int {\frac{{{{\sin }^{ - 1}}x \cdot x}}{{\sqrt {1 - {x^2}} }}} dx\] Let sin^-1x = θ & x = sinθ \[\therefore \frac{1}{{\sqrt {1 - {x^2}} }}dx = d\theta \] \[{I_1} = \int {\theta \sin \theta \cdot d\theta } \] \[I_1 = \theta \int {\sin \theta \cdot d\theta } - \int {\left( {\frac{d}{{dx}}\theta \int {\sin \theta \cdot d\theta } } \right)} d\theta \] \[I_1 = \theta \left( { - \cos \theta } \right) - \int {1\left( { - \cos \theta } \right)} d\theta \] \[{I_1} = - \theta \cos \theta + \sin \theta + c\] \[{I_1} = - \theta \sqrt {1 - {{\sin }^2}\theta } + \sin \theta + c\] \[{I_1} = - \theta \sqrt {1 - {{\sin }^2}\theta } + \sin \theta + c\] substitute value of θ
\[I_1 = - {\sin ^{ - 1}}x\sqrt {1 - {x^2}} + x + c\] Replacing value of I₁ in equation (1)
\[I = x{\left( {{{\sin }^{ - 1}}x} \right)^2} - 2\left[ { - {{\sin }^{ - 1}}x\sqrt {1 - {x^2}} + x} \right] + c\] \[I = x{\left( {{{\sin }^{ - 1}}x} \right)^2} + 2{\sin ^{ - 1}}x\sqrt {1 - {x^2}} - 2x + c\] \[I = x{\left( {{{\sin }^{ - 1}}x} \right)^2} + 2\sqrt {1 - {x^2}} {\sin ^{ - 1}}x - 2x + c\]

Question (11)

\[\frac{{x{{\cos }^{ - 1}}x}}{{\sqrt {1 - {x^2}} }}\]

Solution

\[I = \int {\frac{{x{{\cos }^{ - 1}}x}}{{\sqrt {1 - {x^2}} }}dx} \] Let cos^-1x = θ ⇒ x = cosθ
taking derivative
\[\frac{{ - 1}}{{\sqrt {1 - {x^2}} }}dx = d\theta \] \[\frac{1}{{\sqrt {1 - {x^2}} }}dx = - d\theta \] Replace in given equation
\[I = \int {\cos \theta \cdot \theta \left( { - d\theta } \right)} \] \[I = - \int {\cos \theta \cdot \theta d\theta } \] \[I = - \int {\theta \cdot \cos \theta d\theta } \] \[I = - \left[ {\theta \int {\cos \theta d\theta - \int {\left( {\frac{d}{{d\theta }}\theta \int {\cos \theta d\theta } } \right)d\theta } } } \right]\] \[I = - \left[ {\theta \sin \theta - \int {1 \cdot \sin \theta d\theta } } \right]\] \[I = - \theta \sin \theta + \left( { - \cos \theta } \right) + c\] \[I = - \theta \sqrt {1 - {{\cos }^2}\theta } - \cos \theta + c\] substitute value of θ \[I = - {\cos ^{ - 1}}x\sqrt {1 - {x^2}} - x + c\] \[I = - \left[ {\sqrt {1 - {x^2}} {{\cos }^{ - 1}}x + x} \right] + c\]

Question (12)

xsec²

Solution

\[I = \int {x{{\sec }^2}xdx} \] \[I = x\int {{{\sec }^2}xdx - \int {\left( {\frac{d}{{dx}}x\int {{{\sec }^2}xdx} } \right)} } dx\] \[I = x\tan x - \int {1 \cdot \tan xdx} \] \[I = x\tan x - \left( { - \log \left| {\cos x} \right|} \right) + c\] \[I = x\tan x + \log \left| {\cos x} \right| + c\]

Question (13)

tan^-1x

Solution

\[I = \int {{{\tan }^{ - 1}}xdx} \] \[I = {\tan ^{ - 1}}x\int {1dx - \int {\left( {\frac{d}{{dx}}{{\tan }^{ - 1}}x \cdot \int {1dx} } \right)} } dx\] \[I = {\tan ^{ - 1}}x \cdot x - \int {\frac{1}{{1 + {x^2}}} \cdot } xdx\] \[I = x{\tan ^{ - 1}}x - \int {\frac{x}{{1 + {x^2}}}} dx\] I=x tan^-1 - I₁ equation(1)
\[Here\quad {I_1} = \int {\frac{x}{{1 + {x^2}}}dx} \] Let 1+x² = t
2xdx= dt
xdx= dt/2
\[{I_1} = \int {\frac{{\frac{{dt}}{2}}}{t}} \] \[{I_1} = \frac{1}{2}\int {\frac{{dt}}{t}} \] \[{I_1} = \frac{1}{2}\log \left| t \right| + c\] \[{I_1} = \frac{1}{2}\left| {1 + {x^2}} \right| + c\] Replacing values in equation(1)
\[I = x{\tan ^{ - 1}}x - \frac{1}{2}\log \left| {1 + {x^2}} \right| + c\]

Question (14)

x(logx)²

Solution

\[I = \int {x{{\left( {\log x} \right)}^2}dx} \] \[I = \int {{{\left( {\log x} \right)}^2} \cdot xdx} \] \[I = {\left( {\log x} \right)^2}\int {xdx - \int {\left( {\frac{d}{{dx}}{{\left( {\log x} \right)}^2}\int {xdx} } \right)} } dx\] \[I = {\left( {\log x} \right)^2}\frac{{{x^2}}}{2} - \int {\cancel{2}\left( {\log x} \right) \cdot \frac{1}{x} \cdot \frac{{{x^2}}}{\cancel{2}}dx} \] \[I = \frac{{{{\left( {\log x} \right)}^2}{x^2}}}{2} - \int {\log x \cdot xdx} \] \[Let\quad I = \frac{{{{\left( {\log x} \right)}^2}{x^2}}}{2} - {I_1}\qquad equation(1)\] \[Here\quad {I_1} = \int {\log x \cdot xdx} \] Agian using parts we get \[{I_1} = \log x\int {xdx - \int {\left( {\frac{d}{{dx}}\log x\int {xdx} } \right)dx} } \] \[{I_1} = \log x\frac{{{x^2}}}{2} - \int {\frac{1}{\cancel{x}} \cdot \frac{{{x^\cancel{2}}}}{2}dx} \] \[{I_1} = \frac{{{x^2}\log x}}{2} - \frac{1}{2} \cdot \frac{{{x^2}}}{2} + c\] \[{I_1} = \frac{{{x^2}\log x}}{2} - \frac{{{x^2}}}{4} + c\] Substitute value of I₁ n equation (1)
\[I = \frac{{{x^2}{{\left( {\log x} \right)}^2}}}{2} - \frac{{{x^2}\log x}}{2} + \frac{{{x^2}}}{4} + c\]

Question (15)

(x²+1)logx

Solution

\[I = \int {\left( {{x^2} + 1} \right)\log xdx} \] \[I = \int {\log x \cdot \left( {{x^2} + 1} \right)dx} \] \[I = \log x\int {\left( {{x^2} + 1} \right)dx - \int {\left( {\frac{d}{{dx}}\log x\int {\left( {{x^2} + 1} \right)dx} } \right)} } dx\] \[I = \log x\left( {\frac{{{x^3}}}{3} + x} \right) - \int {\frac{1}{x}\left( {\frac{{{x^3}}}{3} + x} \right)} dx\] \[I = \left( {\frac{{{x^3}}}{3} + x} \right)\log x - \frac{1}{3}\int {\frac{1}{\cancel{x}} \cdot \cancel{x}\left( {{x^2} + 3} \right)} dx\] \[I = \left( {\frac{{{x^3}}}{3} + x} \right)\log x - \frac{1}{3}\int {{x^2}dx - \frac{3}{3}\int {dx} } \] \[I = \left( {\frac{{{x^3}}}{3} + x} \right)\log x - \frac{1}{3}\frac{{{x^3}}}{3} - x + c\] \[I = \left( {\frac{{{x^3}}}{3} + x} \right)\log x - \frac{1}{9}{x^3} - x + c\]

Question (16)

e^x(sinx + cosx)

Solution

\[I = \int {{e^x}} \left( {\sin x + \cos x} \right)dx\] $$I = \int {{e^x}} \left( {\sin x + \cos x} \right)dx$$ Let sin x = f(x)
then f^'(x) = cos x
\[I = \int {{e^x}} \left[ {\sin x + \cos x} \right]dx\] \[I = \int {{e^x}} \left[ {f\left( x \right) + {f'}\left( x \right)} \right]dx\] \[I = {e^x}f\left( x \right) + c\] \[I = {e^x}\sin x + c\]

Question (17)

\[\frac{{x{e^x}}}{{{{\left( {1 + x} \right)}^2}}}\]

Solution

\[I = \int {\frac{{x{e^x}}}{{{{\left( {1 + x} \right)}^2}}}} dx\] \[I = \int {{e^x}} \left[ {\frac{x}{{{{\left( {1 + x} \right)}^2}}}} \right]dx\] \[I = \int {{e^x}} \left[ {\frac{{1 + x - 1}}{{{{\left( {1 + x} \right)}^2}}}} \right]dx\] \[I = \int {{e^x}\left[ {\frac{{\left(\require{cancel}\cancel{ {1 + x}} \right)}}{{{{\left( {1 + x} \right)}^\cancel{2}}}} - \frac{1}{{{{\left( {1 + x} \right)}^2}}}} \right]} dx\] \[I = \int {{e^x}\left[ {\frac{1}{{\left( {1 + x} \right)}} - \frac{1}{{{{\left( {1 + x} \right)}^2}}}} \right]} dx\] \[{\rm{Let }}\;f\left( x \right) = \frac{1}{{1 + x}}\] \[\therefore \quad f'\left( x \right) = \frac{{ - 1}}{{{{\left( {1 + x} \right)}^2}}}\] \[I = \int {{e^x}\left[ {\frac{1}{{1 + x}} + \frac{{ - 1}}{{{{\left( {1 + x} \right)}^2}}}} \right]} dx\] \[I = \int {e\left[ {f\left( x \right) + f'\left( x \right)} \right]} dx\] \[I = {e^x}f\left( x \right) + c\] \[I = {e^x}\frac{1}{{1 + x}} + c\] \[I = \frac{{{e^x}}}{{1 + x}} + c\]

Question (18)

\[{e^x}\frac{{1 + \sin x}}{{1 + \cos x}}\]

Solution

\[I = \int {{e^x}\frac{{1 + \sin x}}{{1 + \cos x}}dx} \] \[I = \int {{e^x}\left( {\frac{1}{{2{{\cos }^2}\frac{x}{2}}} + \frac{{\sin x}}{{2{{\cos }^2}\frac{x}{2}}}} \right)} dx\] \[I = \int {{e^x}\left( {\frac{{{{\sec }^2}\frac{x}{2}}}{2} + \frac{{2\sin \frac{x}{2}\cos \frac{x}{2}}}{{2{{\cos }^2}\frac{x}{2}}}} \right)} dx\] \[I = \int {{e^x}\left[ {\frac{1}{2}{{\sec }^2}\frac{x}{2} + \frac{{\sin \frac{x}{2}}}{{\cos \frac{x}{2}}}} \right]} dx\] \[I = \int {{e^x}\left( {\frac{1}{2}{{\sec }^2}\frac{x}{2} + \tan \frac{x}{2}} \right)} dx\] \[{\rm{Let}}\quad {\rm{f}}\left( x \right) = \tan \frac{x}{2}\] \[f'\left( x \right) = {\sec ^2}\frac{x}{2} \cdot \frac{1}{2}\] \[I = \int {{e^x}\left[ {\frac{1}{2}{{\sec }^2}\frac{x}{2} + \tan \frac{x}{2}} \right]} dx\] \[I = \int {{e^x}\left[ {f'\left( x \right) + f\left( x \right)} \right]} dx\] \[I = {e^x}f\left( x \right) + c\] \[I = {e^x}\tan \frac{x}{2} + c\]

Question (19)

\[{e^x}\left( {\frac{1}{x} - \frac{1}{{{x^2}}}} \right)\]

Solution

\[{\rm{Let f}}\left( {\rm{x}} \right) = \frac{1}{x}\] \[f'\left( x \right) = \frac{{ - 1}}{{{x^2}}}\] \[\int {{e^x}\left( {\frac{1}{x} - \frac{1}{{{x^2}}}} \right)dx} \] \[I = \int {{e^x}} \left[ {f\left( x \right) + f'\left( x \right)} \right]dx\] \[I = {e^x}f\left( x \right) + c\] \[I = {e^x}\frac{1}{x} + c\] \[I = \frac{{{e^x}}}{x} + c\]

Question (20)

\[\frac{{\left( {x - 3} \right){e^x}}}{{{{\left( {x - 1} \right)}^3}}}\]

Solution

\[I = \int {\frac{{\left( {x - 3} \right){e^x}}}{{{{\left( {x - 1} \right)}^3}}}dx} \] \[\frac{{\left( {x - 3} \right)}}{{{{\left( {x - 1} \right)}^3}}} = \frac{{x - 1 - 2}}{{{{\left( {x - 1} \right)}^3}}}\] Now \[I = \int {\frac{{\left( {x - 3} \right){e^x}}}{{{{\left( {x - 1} \right)}^3}}}dx} \] \[I = \int {{e^x}\left[ {\frac{{x - 1 - 2}}{{{{\left( {x - 1} \right)}^3}}}} \right]dx} \] \[I = \int {{e^x}\left[ {\frac{{x - 1}}{{{{\left( {x - 1} \right)}^3}}} - \frac{2}{{{{\left( {x - 1} \right)}^3}}}} \right]} dx\] \[I = \int {{e^x}} \left[ {\frac{1}{{{{\left( {x - 1} \right)}^2}}} - \frac{2}{{{{\left( {x - 1} \right)}^3}}}} \right]dx\] \[\text{let}\quad f\left( x \right) = \frac{1}{{{{\left( {x - 1} \right)}^2}}} = {\left( {x - 1} \right)^{-2}}\] \[f'\left( x \right) = - 2{\left( {x - 1} \right)^{ - 3}}\] \[f'\left( x \right) = \frac{{ - 2}}{{{{\left( {x - 1} \right)}^3}}}\] \[\text{So}\quad I = \int {{e^x}\left[ {\frac{1}{{{{\left( {x - 1} \right)}^2}}} + \frac{{ - 2}}{{{{\left( {x - 1} \right)}^3}}}} \right]} dx\] \[I = \int {{e^x}\left[ {f\left( x \right) + f'\left( x \right)} \right]} dx\] \[I = {e^x}f\left( x \right) + c\] \[I = \frac{{{e^x}}}{{{{\left( {x - 1} \right)}^2}}} + c\]

Question (21)

\[{e^{2x}}\sin x\]

Solution

\[I = \int {{e^{2x}}\sin x\;dx} \] \[I = \int {\sin x\;{e^{2x}}\;dx} \] \[I = \sin x\int {{e^{2x}}dx - \int {\left( {\frac{d}{{dx}}\sin x\int {{e^{2x}}dx} } \right)} } dx\] \[I = \frac{{\sin x{e^{2x}}}}{2} - \int {\cos x} \;\frac{{{e^{2x}}}}{2}dx\] \[I = \frac{{\sin x{e^{2x}}}}{2} - \frac{1}{2}\int {\cos x} \;{e^{2x}}dx\] \[{I_1} = \left[ {\cos x\int {{e^{2x}}dx - \int {\left( {\frac{d}{{dx}}cox\int {{e^{2x}}dx} } \right)dx} } } \right]\] \[{I_1} = \cos x\cdot\frac{{{e^{2x}}}}{2} - \int {\left( { - sinx} \right)} \frac{{{e^{2x}}}}{2}dx\] \[{I_1} = \frac{{\cos x\cdot{e^{2x}}}}{2} + \frac{1}{2}\int {\sin x\cdot{e^{2x}}dx} \] \[I = \frac{{\sin x{e^{2x}}}}{2} - \frac{1}{2}\left[ {{I_1}} \right]\] \[I = \frac{{\sin x{e^{2x}}}}{2} - \frac{1}{2}\left[ {\frac{{\cos x\cdot{e^{2x}}}}{2} + \frac{1}{2}\int {\sin x\cdot{e^{2x}}dx} } \right]\] Here ∫sinx e^2x dx is again initial question can be replaced by I \[I = \frac{{\sin x{e^{2x}}}}{2} - \frac{1}{2}\left[ {\frac{{\cos x\cdot{e^{2x}}}}{2} + \frac{1}{2}I} \right] + c\] \[I = \frac{{\sin x{e^{2x}}}}{2} - \frac{{\cos x\cdot{e^{2x}}}}{4} - \frac{1}{4}I + c\] \[I + \frac{1}{4}I = \frac{{\sin x{e^{2x}}}}{2} - \frac{{\cos x\cdot{e^{2x}}}}{4} + c\] \[\frac{5}{4}I = \frac{{\sin x{e^{2x}}}}{2} - \frac{{\cos x\cdot{e^{2x}}}}{4} + c\] \[I = \frac{\cancel{4}}{5}\left[ {\frac{1}{\cancel{4}}\left( {2\sin x - \cos x} \right){e^{2x}}} \right] + c\] \[I = \frac{{{e^{2x}}}}{5}\left( {2\sin x - \cos x} \right) + c\]

Question (22)

\[{\sin ^{ - 1}}\left( {\frac{{2x}}{{1 + {x^2}}}} \right)\]

Solution

\[I = \int {{{\sin }^{ - 1}}\left( {\frac{{2x}}{{1 + {x^2}}}} \right)dx} \] Let x = tanθ ⇒ θ = tan^-1x
∴ dx = sec²θ dθ
Replacing the value we get \[I = \int {{{\sin }^{ - 1}}\left( {\frac{{2\tan \theta }}{{1 + {{\tan }^2}\theta }}} \right)} \cdot {\sec ^2}\theta d\theta \] \[I = \int {{{\sin }^{ - 1}}\left( {\sin 2\theta } \right)} \cdot {\sec ^2}\theta d\theta \] \[I = \int {2\theta } {\sec ^2}\theta d\theta \] \[I = 2\int \theta {\sec ^2}\theta d\theta \] \[I = 2\left[ {\theta \int {{{\sec }^2}\theta d\theta - \int {\left( {\frac{d}{{d\theta }}\theta \int {{{\sec }^2}\theta d\theta } } \right)d\theta } } } \right]\] \[I = 2\left[ {\theta \tan \theta - \int {1 \cdot \tan \theta d\theta } } \right]\] \[I = 2\theta \tan \theta - 2\log \left| {\sec \theta } \right| + c\] \[I = 2x{\tan ^{ - 1}}x - 2\log \left| {\sqrt {1 + {{\tan }^2}\theta } } \right| + c\] \[I = 2x{\tan ^{ - 1}}x - 2\log \left| {\sqrt {1 + {x^2}} } \right| + c\] \[I = 2x{\tan ^{ - 1}}x - 2\log {\left( {1 + {x^2}} \right)^{\frac{1}{2}}} + c\] \[I = 2x{\tan ^{ - 1}}x - \cancel{2} \cdot \frac{1}{\cancel{2}}\log \left( {1 +{x^2}} \right) + c\] \[I = 2x{\tan ^{ - 1}}x - \log \left( {1 + {x^2}} \right) + c\]

Choose the correct answer in Exercise 23 and 24

Question (23)

\[\int {{x^2}{e^{{x^3}}}} dx\qquad \text{equals}\] Options
\[\left( A \right)\quad \frac{1}{3}{e^{{x^3}}} + c\] \[\left( B \right)\quad\frac{1}{3}{e^{{x^2}}} + c\] \[\left( C \right)\quad\frac{1}{2}{e^{{x^3}}} + c\] \[\left( D \right)\quad\frac{1}{2}{e^{{x^2}}} + c\]

Solution

\[I = \int {{x^2}{e^{{x^3}}}dx} \] \[I = \int {{e^{{x^3}}}{x^2}dx} \] Let x³ = t
3x²dx = dt
x²dx = dt/3
\[I = \int {{e^t}} \frac{{dt}}{3}\] \[I = \frac{1}{3}\int {{e^t}dt} \] \[I = \frac{1}{3}{e^t} + c\] \[I = \frac{1}{3}{e^{{x^3}}} + c\] A is correct option

Question (24)

\[\int {{e^x}\sec x\left( {1 + \tan x} \right)} dx\] Options
(A) e^x cosx + c
(B) e^x secx + c
(C) e^x sinx + c
(D) e^x tanx + c

Solution

\[I = \int {{e^x}\sec x} \left( {1 + \tan x} \right)dx\] \[I = \int {{e^x}\left( {\sec x + \sec x\tan x} \right)} dx\] Let f(x) = secx f'(x) = secx tanx
\[I = \int {{e^x}} \left[ {f\left( x \right) + f'\left( x \right)} \right]dx\] \[I = {e^x}f\left( x \right) + c\] \[I = {e^x}\sec x + c\] B is correct option