Indefinite Integration MCQ Questions & Answers in Calculus | Maths

$$\eqalign{ & I = \int {\frac{1}{{2p}}\sqrt {\frac{{p - 1}}{{p + 1}}} dp} \cr & = \frac{1}{2}\int {\frac{{p - 1}}{{{p\sqrt {\left( {p + 1} \right)\left( {p - 1} \right)} }}} dp} \cr & = \frac{1}{2}\int {\frac{{pdp}}{{p\sqrt {{p^2} - 1} }} - \frac{1}{2}\int {\frac{{dp}}{{p\sqrt {{p^2} - 1} }}} } \cr & = \frac{1}{2}\int {\frac{{dp}}{{\sqrt {{p^2} - 1} }} - \frac{1}{2}\int {\frac{{dp}}{{p\sqrt {{p^2} - 1} }}} } \cr & = \frac{1}{2}{\log _e}\left( {p + \sqrt {{p^2} - 1} } \right) - \frac{1}{2}{\sec ^{ - 1}}p \cr & \Rightarrow f\left( p \right) = \log \sqrt {p + \sqrt {{p^2} - 1} } - \frac{1}{2}{\sec ^{ - 1}}p \cr} $$

$$\eqalign{ & {\text{Let }}I = \int {\frac{{{x^2} - x + 1}}{{{x^2} + 1}}} .{e^{{{\cot }^{ - 1}}x}}dx \cr & {\text{Put }}x = \cot \,t \Rightarrow - {\text{cosec }}t\,dt = dx \cr & {\text{Now, }}1 + {\cot ^2} = {\text{cose}}{{\text{c}}^2}t \cr & \therefore \,I = \int {\frac{{{e^t}\left( {{{\cot }^2}t - \cot \,t + 1} \right)}}{{\left( {{\text{1}} + {{\cot }^2}t} \right)}}} \left( { - {\text{cose}}{{\text{c}}^2}t} \right)dt \cr & = \int {{e^t}\left( {\cot \,t - {\text{cose}}{{\text{c}}^2}t} \right)dt} \cr & = {e^t}\cot \,t + C \cr & = {e^{{{\cot }^{ - 1}}x}}\left( x \right) + C \equiv A\left( x \right).{e^{{{\cot }^{ - 1}}x}} + C \cr & \Rightarrow A\left( x \right) = x \cr} $$

$$\int {\frac{{2{x^{12}} + 5{x^9}}}{{{{\left( {{x^5} + {x^3} + 1} \right)}^3}}}dx} $$

Dividing by $${{x^{15}}}$$ in numerator and denominator

$$\eqalign{ & \int {\frac{{\frac{2}{{{x^3}}} + \frac{5}{{{x^6}}}dx}}{{{{\left( {1 + \frac{1}{{{x^2}}} + \frac{1}{{{x^5}}}} \right)}^3}}}} \cr & {\text{Substitute }}1 + \frac{1}{{{x^2}}} + \frac{1}{{{x^5}}} = t \cr & \Rightarrow \left( {\frac{{ - 2}}{{{x^3}}} - \frac{5}{{{x^6}}}} \right)dx = dt \cr & \Rightarrow \left( {\frac{2}{{{x^3}}} + \frac{5}{{{x^6}}}} \right)dx = - dt \cr & {\text{This gives,}} \cr & \int {\frac{{\frac{2}{{{x^3}}} + \frac{5}{{{x^6}}}dx}}{{{{\left( {1 + \frac{1}{{{x^2}}} + \frac{1}{{{x^5}}}} \right)}^3}}}} = \int {\frac{{ - dt}}{{{t^3}}}} = \frac{1}{{2{t^2}}} + C \cr & = \frac{1}{{2{{\left( {1 + \frac{1}{{{x^2}}} + \frac{1}{{{x^5}}}} \right)}^2}}} + C \cr & = \frac{{{x^{10}}}}{{2{{\left( {{x^5} + {x^3} + 1} \right)}^2}}} + C \cr} $$

$$\eqalign{ & {\text{Let }}\int {f\left( x \right)dx = \psi \left( x \right)} \cr & {\text{Let}}\,I = \int {{x^5}f\left( {{x^3}} \right)dx} \cr & {\text{put }}{x^3} = t\,\, \Rightarrow 3{x^2}dx = dt \cr & I = \frac{1}{3}\int {3.{x^2}.{x^3}.f\left( {{x^3}} \right)} dx \cr & = \frac{1}{3}\int {tf\left( t \right)dt} \cr & = \frac{1}{3}\left[ {t\int {f\left( t \right)dt - } \int {f\left( t \right)dt} } \right] \cr & = \frac{1}{3}\left[ {t\psi \left( t \right) - \int {\psi \left( t \right)dt} } \right] \cr & = \frac{1}{3}\left[ {{x^3}\psi \left( {{x^3}} \right) - 3\int {{x^2}\psi \left( {{x^3}} \right)dx} } \right] + C \cr & = \frac{1}{3}{x^3}\psi \left( {{x^3}} \right) - \int {{x^2}\psi \left( {{x^3}} \right)dx + C} \cr} $$

$$\eqalign{ & {\text{Put }}x = \cos \,2\theta \cr & \therefore I = \int {\cos \left\{ {2{{\tan }^{ - 1}}\tan \,\theta } \right\}} \left( { - 2\sin \,2\theta } \right)d\theta \cr & = - \int {\sin \,4\theta\,d\theta = \frac{1}{4}\cos \,4\theta + c} \cr & = \frac{1}{4}\left( {2{x^2} - 1} \right) + c \cr & = \frac{1}{2}{x^2} + k \cr} $$

$$\eqalign{ & f\left( x \right) = \int {{e^x}\left( {x - 1} \right)\left( {x - 2} \right)dx} \cr & {\text{For decreasing function, }}f'\left( x \right) < 0 \cr & \Rightarrow {e^x}\left( {x - 1} \right)\left( {x - 2} \right) < 0 \cr & \Rightarrow \left( {x - 1} \right)\left( {x - 2} \right) < 0 \cr & \Rightarrow 1 < x < 2 \cr & \therefore \,{e^x} > 0\,\forall \,x\, \in R \cr} $$

$$\eqalign{ & {\text{Let }}I = \int {32{x^3}{{\left( {\log \,x} \right)}^2}dx} \cr & = 32\left\{ {{{\left( {\log \,x} \right)}^2}\frac{{{x^4}}}{4} - \int {2\,\log \,x\frac{1}{x}.\frac{{{x^4}}}{4}dx} } \right\} \cr & = \frac{{32}}{4}{x^4}{\left( {\log \,x} \right)^2} - 16\int {{x^3}\log } \,x\,dx \cr & = 8{x^4}{\left( {\log \,x} \right)^2} - 4{x^4}\log \,x + 4\int {{x^3}dx} \cr & = {x^4}\left\{ {8{{\left( {\log \,x} \right)}^2} - 4\,\log \,x + 1} \right\} + C \cr} $$

$$\eqalign{ & \int {\frac{{\sin \,x}}{{\sin \left( {x - \alpha } \right)}}dx} = \int {\frac{{\sin \,\left( {x - \alpha + \alpha } \right)}}{{\sin \left( {x - \alpha } \right)}}dx} \cr & = \int {\frac{{\sin \left( {x - \alpha } \right)\cos \,\alpha + \cos \left( {x - \alpha } \right)\sin \,\alpha }}{{\sin \left( {x - \alpha } \right)}}dx} \cr & = \int {\left\{ {\cos \,\alpha + \sin \,\alpha \,\cot \left( {x - \alpha } \right)} \right\}dx} \cr & = \left( {\cos \,\alpha } \right)x + \left( {\sin \,\alpha } \right)\log \,\sin \,\left( {x - \alpha } \right) + C \cr & \therefore A = \cos \,\alpha ,\,\,\,B = \sin \,\alpha \cr} $$

$$\eqalign{ & \int {\frac{{5\,\tan \,x}}{{\tan \,x - 2}}dx} = \int {\frac{{5\frac{{\sin \,x}}{{\cos \,x}}}}{{\frac{{\sin \,x}}{{\cos \,x}} - 2}}dx} \cr & = \int {\left( {\frac{{5\sin \,x}}{{\cos \,x}} \times \frac{{\cos \,x}}{{\sin \,x - 2\,\cos \,x}}} \right)dx} \cr & = \int {\frac{{5\sin \,x\,dx}}{{\sin \,x - 2\cos \,x}}} \cr & = \int {\left( {\frac{{4\sin \,x + \sin \,x + 2\cos \,x - 2\cos \,x}}{{\sin \,x - 2\cos \,x}}} \right)} dx \cr & = \int {\left( {\frac{{\left( {\sin \,x - 2\cos \,x} \right) + \left( {4sin\,x + 2\cos \,x} \right)}}{{\sin \,x - 2\cos \,x}}} \right)} dx \cr & = \int {\left( {\frac{{\left( {\sin \,x - 2\cos \,x} \right) + 2\left( {\cos \,x + 2\sin \,x} \right)}}{{\sin \,x - 2\cos \,x}}} \right)} dx \cr & = \int {\frac{{\sin \,x - 2\,\cos \,x}}{{\sin \,x - 2\cos \,x}}} dx + 2\int {\left( {\frac{{\cos \,x + 2\sin \,x}}{{\sin \,x - 2\cos \,x}}} \right)dx} \cr & = \int {dx + 2} \int {\frac{{\cos \,x + 2\sin \,x}}{{\sin \,x - 2\cos \,x}}dx} \cr & = {I_1} + {I_2} \cr & {\text{Where}}\,\,{I_1} = \int {dx} {\text{ and}} \cr & {I_2} = 2\int {\frac{{\cos \,x + 2\sin \,x}}{{\sin \,x - 2\cos \,x}}} dx \cr & {\text{Put }}\sin \,x - 2\cos \,x = t \cr & \Rightarrow \left( {\cos \,x + 2\sin \,x} \right)dx = dt \cr & \therefore {I_2} = 2\int {\frac{{dt}}{t} = 2\ln \,t + C = 2\,\ln \,} \left( {\sin \,x - 2\cos \,x} \right) + C \cr & {\text{Hence,}}\,\,{I_1} + {I_2} = \int {dx + 2\ln \left( {\sin \,x - 2\cos \,x} \right) + c} \cr & = x + 2\ln \left|( {\sin \,x - 2\cos \,x} \right)| + k\,\,\, \Rightarrow a = 2 \cr} $$

$$\eqalign{ & I = \int\limits_{ - \frac{\pi }{2}}^{\frac{\pi }{2}} {\frac{{\ln \left( {\cos \,x} \right)}}{{1 + {e^x}.{e^{\sin \,x}}}}dx} \cr & \Rightarrow I = \int\limits_{ - \frac{\pi }{2}}^{\frac{\pi }{2}} {\frac{{\ln \left( {\cos \,x} \right)}}{{1 + {e^{ - \left( {x + \sin \,x} \right)}}}}dx} \cr & \Rightarrow 2I = \int\limits_{ - \frac{\pi }{2}}^{\frac{\pi }{2}} {\frac{{\ln \left( {\cos \,x} \right)}}{{1 + {e^{x + \sin \,x}}}}\left( {1 + {e^{\left( {x + \sin \,x} \right)}}} \right)} dx \cr & \Rightarrow 2I = \int\limits_{ - \frac{\pi }{2}}^{\frac{\pi }{2}} {\ln \left( {\cos \,x} \right)dx} \cr & \Rightarrow 2I = 2\int\limits_0^{\frac{\pi }{2}} {\ln \left( {\cos \,x} \right)dx} \cr & \Rightarrow I = - \frac{\pi }{2}\ln \,2 \cr} $$