Indefinite Integration MCQ Questions & Answers in Calculus | Maths

$$\eqalign{ & {\text{Let }}I = \int {\frac{{{e^x}\left( {1 + x} \right)}}{{{{\cos }^2}\left( {x{e^x}} \right)}}dx} \cr & {\text{Put, }}x{e^x} = t \Rightarrow {e^x}\left( {1 + x} \right)dx = dt \cr & \therefore \,I = \int {\frac{{dt}}{{{{\cos }^2}t}}} = \int {{{\sec }^2}tdt = } \tan \,t + c = \tan \left( {x{e^x}} \right) + c \cr} $$
where $$'c'$$ is a constant of integration.

Putting $${l^{r + 1}}\left( x \right) = t$$   and $$\frac{1}{{xl\left( x \right){l^2}\left( x \right).....{l^r}\left( x \right)}}dx = dt$$
$$\eqalign{ & {\text{we get, }}\int {\frac{1}{{x{l^2}\left( x \right){l^3}\left( x \right).....{l^r}\left( x \right)}}} \cr & = \int {1.dt} \cr & = t + C \cr & = {l^{r + 1}}\left( x \right) + C \cr} $$

$$\eqalign{ & {\text{If,}} \cr & I = \int {{{\sin }^{ - \frac{{11}}{3}}}x\,{{\cos }^{ - \frac{1}{3}}}x\,dx} , \cr & {\text{here }}\frac{{ - \frac{{11}}{3} - \frac{1}{3} - 2}}{2} = - 3\,\,\left( {{\text{a negative integer}}} \right) \cr & I = \int {\frac{{{{\sin }^{ - \frac{{11}}{3}}}x}}{{{{\cos }^{ - \frac{{11}}{3}}}x}}} {\cos ^{ - \frac{1}{3}}}x.{\cos ^{ - \frac{{11}}{3}}}x\,dx \cr & I = \int {{{\left( {\tan \,x} \right)}^{ - \frac{{11}}{3}}}x.{{\left( {\cos \,x} \right)}^{ - 4}}dx} \cr & I = \int {{{\left( {\tan \,x} \right)}^{ - \frac{{11}}{3}}}x.{{\sec }^4}xdx} \cr & I = \int {{{\left( {\tan \,x} \right)}^{ - \frac{{11}}{3}}}\left( {1 + {{\tan }^2}\,x} \right){{\sec }^2}x\,dx} \cr & {\text{Put }}\tan \,x = t,\,{\sec ^2}x\,dx = dt \cr & I = \int {{t^{ - \frac{{11}}{3}}}} \left( {1 + {t^2}} \right)dt \cr & I = \int {\left( {{t^{ - \frac{{11}}{3}}} + {t^{ - \frac{5}{3}}}} \right)} dt \cr & I = \frac{{{t^{ - \frac{8}{3}}}}}{{ - \frac{8}{3}}} + \frac{{{t^{ - \frac{2}{3}}}}}{{ - \frac{2}{3}}} + C \cr & I = - \frac{3}{8}{\left( {\tan \,x} \right)^{ - \frac{8}{3}}} - \frac{3}{2}{\left( {\tan \,x} \right)^{ - \frac{2}{3}}} + C \cr & I = - \frac{3}{2}{\cot ^{\frac{2}{3}}}x - \frac{3}{8}{\cot ^{\frac{8}{3}}}x + C \cr} $$

$$\eqalign{ & {\text{Let,}}\,\,\,\,\,I = \int {\frac{{{{\cos }^3}x + {{\cos }^5}x}}{{{{\sin }^2}x + {{\sin }^4}x}}dx} \cr & = \int {\frac{{\left( {{{\cos }^2}x + {{\cos }^4}x} \right)\cos \,x}}{{{{\sin }^2}x\left( {1 + {{\sin }^2}x} \right)}}dx} \cr & = \int {\frac{{\left[ {1 - {{\sin }^2}x + {{\left( {1 - {{\sin }^2}x} \right)}^2}} \right]\cos x}}{{{{\sin }^2}x\left( {1 + {{\sin }^2}x} \right)}}dx} \cr & = \int {\frac{{\left( {2 - 3\,{{\sin }^2}x + {{\sin }^4}x} \right)\cos \,x}}{{{{\sin }^2}x\left( {1 + {{\sin }^2}x} \right)}}dx} \cr & {\text{Put }}\sin \,x = t\,\,\, \Rightarrow \cos \,x\,dx = dt \cr & I = \int {\frac{{2 - 3{t^2} + {t^4}}}{{{t^4} + {t^2}}}dt,\,\,\,} I = \int {\left( {1 + \frac{2}{{{t^2}}} - \frac{6}{{{t^2} + 1}}} \right)dt\,} \cr & = t - \frac{2}{t} - 6\,{\tan ^{ - 1}}\left( t \right) + C \cr & = \sin \,x - 2{\left( {\sin \,x} \right)^{ - 1}} - 6\,{\tan ^{ - 1}}\left( {\sin \,x} \right) + C \cr} $$

$$I = \int {\log \frac{{\phi \left( x \right)}}{{f\left( x \right)}}d\left\{ {\log \frac{{\phi \left( x \right)}}{{f\left( x \right)}}} \right\}} = \frac{1}{2}{\left\{ {\log \frac{{\phi \left( x \right)}}{{f\left( x \right)}}} \right\}^2} + k$$

We check from the given options one by one. Options (A) and (B) do not satisfy. We check option (C).
$$\eqalign{ & {\text{Let }}f\left( x \right) = \frac{x}{2} + \alpha \cr & \therefore \,\int {\frac{{dx}}{{\frac{x}{2} + \alpha }}} \cr & = \int {\frac{{2dx}}{{\left( {x + 2\alpha } \right)}}} \cr & = 2\,\log \left( {x + 2\alpha } \right) + {c_1} \cr & = \log {\left( {x + 2\alpha } \right)^2} + {c_1} \cr & = \log {\left( {\frac{x}{2} + \alpha } \right)^2} + \log \,{2^2} + {c_1} \cr & = \log {\left( {\frac{x}{2} + \alpha } \right)^2} + c \cr} $$

$$\eqalign{ & {\text{Given}} \cr & I = \int {\frac{{{e^x}}}{{{e^{4x}} + {e^{2x}} + 1}}dx,\,\,\,\,} \cr & J = \int {\frac{{{e^{ - \,x}}}}{{{e^{ - \,4x}} + {e^{ - \,2x}} + 1}}dx} = \int {\frac{{{e^{3x}}}}{{{e^{4x}} + {e^{2x}} + 1}}dx} \cr & \therefore J - I = \int {\frac{{{e^x}\left( {{e^{2x}} - 1} \right)}}{{{e^{4x}} + {e^{2x}} + 1}}dx} \cr & {\text{Let }}{e^x} = t\,\, \Rightarrow {e^x}\,dx = dt \cr & \therefore J - I\int {\frac{{{t^2} - 1}}{{{t^4} + {t^2} + 1}}} dt = \int {\frac{{1 - \frac{1}{{{t^2}}}}}{{{t^2} + 1 + \frac{1}{{{t^2}}}}}dt} \cr & {\text{Let }}t - \frac{1}{t} = u\,\, \Rightarrow \left( {1 - \frac{1}{{{t^2}}}} \right)dt = du \cr & \therefore J - I = \int {\frac{{du}}{{{u^2} - 1}} = \frac{1}{2}\log \left| {\frac{{u - 1}}{{u + 1}}} \right| + C} \cr & = \frac{1}{2}\log \left| {\frac{{\frac{{{t^2} + 1}}{t} - 1}}{{\frac{{{t^2} + 1}}{t} + 1}}} \right| + C \cr & = \frac{1}{2}\log \left| {\frac{{{e^{2x}} - {e^x} + 1}}{{{e^{2x}} + {e^x} + 1}}} \right| + C \cr} $$

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

$$\eqalign{ & {\text{Put }}x = {z^4}\, \Rightarrow dx = 4{z^3}dz \cr & \therefore \,\int {\frac{{\root 2 \of x }}{{1 + \root 4 \of {{x^3}} }}dx} \cr & = \int {\frac{{{z^2}.4{z^3}}}{{1 + {z^3}}}dz} \cr & = 4\int {\frac{{{z^3}.{z^2}}}{{{z^3} + 1}}dz} \cr & = \frac{4}{3}\int {\frac{{\left( {y - 1} \right)}}{y}dy\,\,\,\,\,\,} \left[ {{\text{Putting }}{z^3} + 1 = y \Rightarrow {z^2}dz = \frac{1}{3}dy} \right] \cr & = \frac{4}{3}\left( {y - \log \,y} \right) + C \cr & = \frac{4}{3}\left[ {1 + {x^{\frac{3}{4}}} - \log \left( {1 + {x^{\frac{3}{4}}}} \right)} \right] + C \cr} $$

$$I = \int {{{\log }_e}\left( {\sqrt {1 - x} + \sqrt {1 + x} } \right)} .1\,dx$$
Integrating by parts taking 1 as the second function.
$$\eqalign{ & I = \log \left( {\sqrt {1 - x} + \sqrt {1 + x} } \right)x - \int {\frac{1}{{\sqrt {1 - x} + \sqrt {1 + x} }}\left[ { - \frac{1}{{2\sqrt {1 - x} }} + \frac{1}{{2\sqrt {1 + x} }}} \right]\left( x \right)dx} \cr & \,\,\,\,\, = x\,\log \left( {\sqrt {1 - x} + \sqrt {1 + x} } \right) - \frac{1}{2}\int {\frac{{\sqrt {1 - x} - \sqrt {1 + x} }}{{\sqrt {1 - x} + \sqrt {1 + x} }}.\frac{1}{{\sqrt {1 - {x^2}} }}.x\,dx} \cr & \,\,\,\,\, = x\,\log \left( {\sqrt {1 - x} + \sqrt {1 + x} } \right) - \frac{1}{2}\int {\frac{{\left( {1 - x} \right) + \left( {1 + x} \right) - 2\sqrt {1 - {x^2}} }}{{\left( {1 - x} \right) - \left( {1 + x} \right)}}} .\frac{1}{{\sqrt {1 - {x^2}} }}.x\,dx \cr & \,\,\,\,\, = x\,\log \left( {\sqrt {1 - x} + \sqrt {1 + x} } \right) - \frac{1}{2}\int {\frac{{\sqrt {1 - {x^2}} - 1}}{{\sqrt {1 - {x^2}} }}} dx \cr & \,\,\,\,\, = x\,\log \left( {\sqrt {1 - x} + \sqrt {1 + x} } \right) - \frac{1}{2}\left[ {\int {1\,dx - \int {\frac{1}{{\sqrt {1 - {x^2}} }}} } dx} \right] \cr & \,\,\,\,\, = x\,\log \left( {\sqrt {1 - x} + \sqrt {1 + x} } \right) + \frac{1}{2}\left[ {{{\sin }^{ - 1}}x - x} \right] + C \cr & \therefore \,f\left( x \right) = x,\,g\left( x \right) = {\sin ^{ - 1}}x - x \cr} $$