Indefinite Integration MCQ Questions & Answers in Calculus | Maths

$$\eqalign{ & \int {{{\cot }^4}x\,dx} \cr & = \int {{{\cot }^2}x.\left( {{\text{cose}}{{\text{c}}^2}x - 1} \right)dx} \cr & = \int {{{\cot }^2}x\,{\text{cose}}{{\text{c}}^2}x\,dx} - \int {\left( {{\text{cose}}{{\text{c}}^2}x - 1} \right)dx} \cr & = - \frac{1}{3}{\cot ^3}x + \cot \,x + x + c \cr & \therefore \,\phi \left( x \right) = - \frac{1}{3}{\cot ^3}x + \cot \,x + x + c + \frac{1}{3}{\cot ^3}x - \cot \,x = x + c \cr & \therefore \,\phi \left( {\frac{\pi }{2}} \right) = \frac{\pi }{2} + c, \cr & \therefore \,c = 0 \cr & \therefore \,\phi \left( x \right) = x \cr} $$

$$\eqalign{ & \int {\frac{{{2^x}}}{{\sqrt {1 - {4^x}} }}dx} \cr & {\text{Put }}{2^x} = t \cr & {2^x}{\log _e}2dx = dt \cr & = \int {\frac{1}{{{{\log }_e}2\sqrt {1 - {t^2}} }}dx} \cr & = {\log _2}e\,{\sin ^{ - 1}}t + k \cr & = {\log _2}e\,{\sin ^{ - 1}}\left( {{2^x}} \right) + k \cr} $$

$$\eqalign{ & {\text{Putting }}x = - z,\,I = \int { - {e^z}\left( {\sec \,z + \sec \,z.\tan \,z} \right)dz} \cr & \therefore I = - \int {{e^z}} \left\{ {\sec \,z + \frac{d}{{dz}}\left( {\sec \,z} \right)} \right\}dz \cr & = - {e^z}\sec \,z + k \cr & = - {e^{ - x}}\sec \,x + k \cr} $$

$$\eqalign{ & {\text{Put }}ln\,x = t \cr & \Rightarrow \frac{1}{x}dx = dt \Rightarrow dx = x\,dt = {e^t}dt \cr & \therefore \,I = \int {\left( {ln\,t + \frac{1}{{{t^2}}}} \right){e^t}dt} \cr & = \int {\left( {ln\,t + \frac{1}{t} - \frac{1}{t} + \frac{1}{{{t^2}}}} \right){e^t}dt} \cr & = \int {\left( {ln\,t + \frac{1}{t}} \right){e^t}dt + } \int {{e^t}\left( { - \frac{1}{t} + \frac{1}{{{t^2}}}} \right)dt} \cr & = {e^t}ln\,t - \frac{{{e^t}}}{t} + c\,\,\,\,\,\left[ {\because \,\frac{d}{{dt}}ln\,t = \frac{1}{t}{\text{ and }}\frac{d}{{dt}}\left( { - \frac{1}{t}} \right) = \frac{1}{{{t^2}}}} \right] \cr & = x\,ln\left( {ln\,x} \right) - \frac{x}{{ln\,x}} + c \cr} $$

$$\eqalign{ & I = \int {\frac{{dx}}{{{x^2}{{\left( {{x^4} + 1} \right)}^{\frac{3}{4}}}}} = \int {\frac{{dx}}{{{x^3}{{\left( {1 + {x^{ - 4}}} \right)}^{\frac{3}{4}}}}}} } \cr & {\text{Let}}\,\,{x^{ - \,4}} = y \cr & \Rightarrow - 4{x^{ - \,3}}dx = dy\,\, \Rightarrow dx = \frac{{ - 1}}{4}{x^3}dy \cr & \therefore I = \frac{{ - 1}}{4}\int {\frac{{{x^3}dy}}{{{x^3}{{\left( {1 + y} \right)}^{\frac{3}{4}}}}}} \cr & = \frac{{ - 1}}{4}\int {\frac{{dy}}{{{{\left( {1 + y} \right)}^{\frac{3}{4}}}}}} \cr & = \frac{{ - 1}}{4} \times 4{\left( {1 + y} \right)^{\frac{1}{4}}} \cr & = - {\left( {1 + {x^{ - \,4}}} \right)^{\frac{1}{4}}} + C \cr & = - {\left( {\frac{{{x^4} + 1}}{{{x^4}}}} \right)^{\frac{1}{4}}} + C \cr} $$

$$\eqalign{ & {\text{Let }}I = \int {{{\tan }^2}x\,{{\sec }^4}x\,dx} \cr & {\text{Let }}\tan \,x = t \cr & \Rightarrow {\sec ^2}x\,dx = dt \cr & \therefore \,I = \int {{{\tan }^2}x.{{\sec }^2}x.{{\sec }^2}x.dx} \cr & = \int {{{\tan }^2}x\left( {1 + {{\tan }^2}x} \right){{\sec }^2}x.dx} \cr & \therefore \,I = \int {{t^2}\left( {1 + {t^2}} \right)dt} \cr & = \int {\left( {{t^2} + {t^4}} \right)dt} \cr & = \frac{{{t^5}}}{5} + \frac{{{t^3}}}{3} + c \cr & = \frac{{{{\tan }^5}x}}{5} + \frac{{{{\tan }^3}x}}{3} + c \cr} $$

$$\eqalign{ & I = \int {\frac{{\left( {1 + x} \right)}}{{x{{\left( {1 + x{e^x}} \right)}^2}}}dx} \cr & \,\,\,\,\, = \int {\frac{{\left( {1 + x} \right){e^x}}}{{x{e^x}{{\left( {1 + x{e^x}} \right)}^2}}}dx} \cr & {\text{Put }}x{e^x} = t \Rightarrow \left( {{e^x} + x{e^x}} \right)dx = dt \cr & I = \int {\frac{{dt}}{{t{{\left( {1 + t} \right)}^2}}}} \cr & {\text{Let }}\frac{1}{{t{{\left( {1 + t} \right)}^2}}} = \frac{A}{t} + \frac{B}{{1 + t}} + \frac{D}{{{\left( {1 + t} \right)}^2}},{\text{ we get}} \cr & A = \frac{1}{{{{\left( {1 + 0} \right)}^2}}} = 1,\,\,D = \frac{1}{{ - 1}} = - 1 \cr & {\text{Equating coefficient of }}{t^2},\,\,\,0 = A + B \Rightarrow B = - 1 \cr & \therefore \,I = \int {\left[ {\frac{1}{t} - \frac{1}{{1 + t}} - \frac{1}{{{{\left( {1 + t} \right)}^2}}}} \right]} dt \cr & \,\,\,\,\,\,\,\,\,\,\,\,\, = \ln \left| t \right| - \ln \left| {1 + t} \right| + \frac{1}{{1 + t}} + C \cr & \,\,\,\,\,\,\,\,\,\,\,\,\, = \ln \left| {\frac{t}{{1 + t}}} \right| + \frac{1}{{1 + t}} + C \cr & \,\,\,\,\,\,\,\,\,\,\,\,\, = \ln \left| {\frac{{x{e^x}}}{{1 + x{e^x}}}} \right| + \frac{1}{{1 + x{e^x}}} + C \cr} $$

$$\eqalign{ & I = \int {\frac{{dx}}{{\cos \,x + \sqrt 3 \sin \,x}}} \cr & \Rightarrow I = \int {\frac{{dx}}{{2\left[ {\frac{1}{2}\cos \,x + \frac{{\sqrt 3 }}{2}\sin \,x} \right]}}} \cr & \Rightarrow I = \frac{1}{2}.\int {\frac{{dx}}{{\left[ {\sin \frac{\pi }{6}\cos \,x + \cos \frac{\pi }{6}\sin \,x} \right]}}} \cr & \Rightarrow I = \frac{1}{2}.\int {\frac{{dx}}{{\sin \left( {x + \frac{\pi }{6}} \right)}}} \cr & \Rightarrow I = \frac{1}{2}.\int {{\text{cosec}}\left( {x + \frac{\pi }{6}} \right)dx} \cr & \because \,\int {{\text{cosec}}\,x\,dx} = \log \left| {\left( {\tan \frac{x}{2}} \right)} \right| + C \cr & \therefore \,I = \frac{1}{2}.\log \,\tan \left( {\frac{x}{2} + \frac{\pi }{{12}}} \right) + C \cr} $$

$$\eqalign{ & y = \int {\frac{1}{3}{e^{{x^3}}}d\left( {{x^3}} \right)} = \frac{1}{3}{e^{{x^3}}} + c \cr & {\text{It passes through }}(0,{\text{ }}1) \cr & \therefore 1 = \frac{1}{3}{e^0} + c \cr & \therefore c = \frac{2}{3}.{\text{ Hence, }}y = \frac{1}{3}\left( {{e^{{x^3}}} + 2} \right) \cr & \therefore {e^{{x^3}}} = 3y - 2\,\,\,\,\,\,\,\,\,\,\therefore x = \root 3 \of {{{\log }_e}\left( {3y - 2} \right)} \cr} $$

$$\eqalign{ & I = \int {\frac{{\sin \,x\,dx}}{{4\,\sin \,x\,\cos \,x\,\cos \,2x}}} \cr & = \frac{1}{4}\int {\frac{{\cos \,x\,dx}}{{\left( {1 - {{\sin }^2}x} \right)\left( {1 - 2\,{{\sin }^2}x} \right)}}} \cr & = \frac{1}{4}\int {\frac{{dt}}{{\left( {1 - {t^2}} \right)\left( {1 - 2{t^2}} \right)}}\,\,\,\,\,\,\,\,\,\,\left[ {t = \sin \,x} \right]} \cr & = \frac{1}{4}\int {\left( {\frac{2}{{1 - 2{t^2}}} - \frac{1}{{1 - {t^2}}}} \right)dt} \cr & = \frac{1}{4}\left\{ {\frac{2}{{2\sqrt 2 }}\log \left| {\frac{{1 + \sqrt 2 t}}{{1 - \sqrt 2 t}}} \right| - \frac{1}{2}\log \left| {\frac{{1 + t}}{{1 - t}}} \right|} \right\} + C \cr & = \frac{1}{8}\log \left| {\frac{{\sin \,x - 1}}{{\sin \,x + 1}}} \right| - \frac{1}{{4\sqrt 2 }}\log \left| {\frac{{\sqrt 2 \,\sin \,x - 1}}{{\sqrt 2 \,\sin \,x + 1}}} \right| + C \cr} $$