Indefinite Integration MCQ Questions & Answers in Calculus | Maths

$$\eqalign{ & I = \int {\frac{{dx}}{{{{\left( {\sin \,x} \right)}^{\frac{4}{3}}}.{{\left( {\cos \,x} \right)}^{\frac{2}{3}}}}}} \cr & \Rightarrow I = \int {\frac{{dx}}{{{{\left( {\frac{{\sin \,x}}{{\cos \,x}}} \right)}^{\frac{4}{3}}}.{\cos^2}\,x}}} \cr & \Rightarrow I = \int {\frac{{{{\sec }^2}x}}{{{{\left( {\tan \,x} \right)}^{\frac{4}{3}}}}}dx} \cr & {\text{Put }}\tan \,x = t \Rightarrow {\sec ^2}x\,dx = dt \cr & \therefore \,I = \int {\frac{{dt}}{{{t^{\frac{4}{3}}}}}} \Rightarrow I = \frac{{ - 3}}{{{t^{\frac{1}{3}}}}} + C \cr & \Rightarrow I = \frac{{ - 3}}{{{{\left( {\tan \,x} \right)}^{\frac{1}{3}}}}} + C \cr & \Rightarrow I = - 3{\mkern 1mu} {\tan ^{ - {\kern 1pt} \frac{1}{3}}}x + C \cr} $$

$$\eqalign{ & \int {f\left( x \right)\sin \,x\,\cos \,x\,dx = \frac{1}{{2\left( {{b^2} - {a^2}} \right)}}{{\log }_e}\left( {f\left( x \right)} \right)} + C \cr & {\text{Therefore, }} \cr & f\left( x \right)\sin \,x\,\cos \,x = \frac{1}{{2\left( {{b^2} - {a^2}} \right)}}.\frac{1}{{f\left( x \right)}}f'\left( x \right)\,\,\,\left[ {{\text{by differentiating both the sides}}} \right] \cr & \Rightarrow 2\left( {{b^2} - {a^2}} \right)\sin \,x\,\cos \,x = \frac{{f'\left( x \right)}}{{{{\left( {f\left( x \right)} \right)}^2}}} \cr & \int {\left( {2{b^2}\sin \,x\,\cos \,x - 2{a^2}\sin \,x\,\cos \,x} \right)dx = } \int {\frac{{f'\left( x \right)}}{{{{\left( {f\left( x \right)} \right)}^2}}}} \,dx \cr & \left[ {{\text{by integrating both the sides}}} \right] \cr & \Rightarrow - {b^2}{\cos ^2}x - {a^2}{\sin ^2}x - C = - \frac{1}{{f\left( x \right)}} \cr} $$

$$\eqalign{ & {\text{Let }}I = \int {\sec \,x.{\text{cosec}}\,x\,dx} \cr & = \int {\frac{2}{{2\,\sin \,x\,\cos \,x}}dx} \cr & = 2\int {\frac{1}{{\sin \,2x}}dx - 2\int {\frac{1}{{\frac{{2\,\tan \,x}}{{1 + {{\tan }^2}x}}}}} } \,\,\,\,\,\left[ {\because \,\sin \,2x = \frac{{2\,\tan \,x}}{{1 + {{\tan }^2}x}}} \right] \cr & = \int {\frac{{{{\sec }^2}x}}{{\tan \,x}}dx} \cr & {\text{Let }}\tan \,x = t \Rightarrow {\sec ^2}dx = dt \cr & {\text{So, }}I = \int {\frac{{dt}}{t}} = \log \left| t \right| + c = \log \left| {\tan \,x} \right| + c \cr & {\text{But}}\,\int {\sec \,x\,{\text{cosec}}\,x\,dx = \log \left| {g\left( x \right)} \right|} + c \cr & \therefore \,g\left( x \right) = \tan \,x \cr} $$

$$\eqalign{ & \int {\sqrt {1 + {x^2}} d\left( {{x^2}} \right)} = \int {\sqrt {1 + t} } \,dt,\,{\text{where }}{x^2} = t \cr & = \frac{{{{\left( {1 + t} \right)}^{\frac{3}{2}}}}}{{\frac{3}{2}}} + k \cr & = \frac{2}{3}{\left( {1 + {x^2}} \right)^{\frac{3}{2}}} + k \cr} $$

$$\eqalign{ & {\text{Since }}\int {{{\left[ {f\left( x \right)} \right]}^n}f'\left( x \right)dx = \frac{{{{\left[ {f\left( x \right)} \right]}^{n + 1}}}}{{n + 1}}} + c \cr & \therefore \,\int {f\left( x \right)\cos \,x\,dx = \frac{{{f^2}\left( x \right)}}{2} + c} \cr & \Rightarrow f'\left( x \right) = \cos \,x \cr & \Rightarrow f\left( x \right) = \sin \,x \cr} $$

$$\eqalign{ & {I_n} = \int {{{\tan }^n}x\,dx,\,n > 1} \cr & {\text{Let }}I = {I_4} + {I_6} \cr & = \int {\left( {{{\tan }^4}x + {{\tan }^6}x} \right)dx} \cr & = \int {{{\tan }^4}x\,\,{{\sec }^2}x\,\,dx} \cr & {\text{Let}}\,\,\,\tan \,x = t \cr & \Rightarrow {\sec ^2}x\,dx = dt \cr & \therefore I = \int {{t^{^4}}dt} = \frac{{{t^5}}}{5} + C \cr & = \frac{1}{5}{\tan ^5}x + C \cr} $$
$$ \Rightarrow $$ On comparing, we have
$$a = \frac{1}{5},\,\,\,\,b = 0$$

$$\eqalign{ & \int {\frac{{{x^2} - 1}}{{{x^3}\sqrt {2{x^4} - 2{x^2} + 1} }}dx} \cr & = \int {\frac{{{x^2} - 1}}{{{x^5}\sqrt {2 - \frac{2}{{{x^2}}} + \frac{1}{{{x^4}}}} }}dx} \cr & = \frac{1}{4}\int {\frac{{\frac{4}{{{x^3}}} - \frac{4}{{{x^5}}}}}{{\sqrt {2 - \frac{2}{{{x^2}}} + \frac{1}{{{x^4}}}} }}dx} \cr & {\text{Put }}2 - \frac{2}{{{x^2}}} + \frac{1}{{{x^4}}} = t \cr & \Rightarrow \left( {\frac{4}{{{x^3}}} - \frac{4}{{{x^5}}}} \right)dx = dt \cr & = \frac{1}{4}\int {\frac{{dt}}{{\sqrt t }} = \frac{{2\sqrt t }}{4} + C} \cr & = \frac{{\sqrt {2 - \frac{2}{{{x^2}}} + \frac{1}{{{x^4}}}} }}{2} + C \cr & = \frac{{\sqrt {2{x^4} - 2{x^2} + 1} }}{{2{x^2}}} + C \cr} $$

$$\eqalign{ & I = \int {{{\left( {x + \frac{1}{x}} \right)}^{n + 5}}\left( {\frac{{{x^2} - 1}}{{{x^2}}}} \right)} dx \cr & {\text{Put }}x + \frac{1}{x} = t \cr & \Rightarrow \left( {1 - \frac{1}{{{x^2}}}} \right)dx = dt \cr & \Rightarrow \left( {\frac{{{x^2} - 1}}{{{x^2}}}} \right)dx = dt \cr & \therefore \,I = \int {{t^{n + 5}}} = \frac{{{t^{n + 6}}}}{{n + 6}} + c = \frac{{{{\left( {x + \frac{1}{x}} \right)}^{n + 6}}}}{{n + 6}} + c \cr} $$

$$\eqalign{ & {\text{Put }}1 + {x^{\frac{4}{5}}} = t \cr & \therefore \frac{4}{5}{x^{ - \,\frac{1}{5}}}dx = dt \cr & \therefore I = \frac{5}{4}\int {\frac{{dt}}{{{t^{\frac{1}{2}}}}}} \cr & = \frac{5}{4}.2.\sqrt t + k \cr & = \frac{5}{2}\sqrt {1 + {x^{\frac{4}{5}}}} + k \cr} $$

$$\eqalign{ & \int {\frac{{{x^2}}}{{\left( {{x^2} + 1} \right)\left( {{x^2} + 4} \right)}}} dx \cr & = \frac{1}{3}\int {\left[ {\frac{4}{{{x^2} + 4}} - \frac{1}{{{x^2} + 1}}} \right]} dx \cr & = - \frac{1}{3}\int {\frac{1}{{{x^2} + 1}}dx} + \frac{4}{3}\int {\frac{1}{{{x^2} + 4}}dx} \cr & = - \frac{1}{3}{\tan ^{ - 1}}x + \frac{4}{3} \times \frac{1}{2}{\tan ^{ - 1}}\left( {\frac{x}{2}} \right) + C \cr & = - \frac{1}{3}{\tan ^{ - 1}}x + \frac{2}{3}{\tan ^{ - 1}}\left( {\frac{x}{2}} \right) + C \cr} $$