Indefinite Integration MCQ Questions & Answers in Calculus | Maths

$$\eqalign{ & I = \int {2\,\sin \,x.\cos \,x.\log \,\cos \,x\,dx} \cr & {\text{Put }}\log \,\cos \,x = t \cr & \therefore \, - \frac{{\sin \,x}}{{\cos \,x}}dx = dt \cr & I = \int {2\,\sin \,x.\cos \,x.t\frac{{\cos \,x}}{{ - \sin \,x}}dt} \cr & = - 2\int {{{\cos }^2}x.t\,dt = - 2\int {t{e^{2t}}dt} } \cr & = - 2\left[ {t.\frac{{{e^{2t}}}}{2} - \int {\frac{{{e^{2t}}}}{2}.dt} } \right] \cr & = - t{e^{2t}} + \frac{1}{2}{e^{2t}} + k \cr & = {e^{2t}}\left( {\frac{1}{2} - t} \right) + k \cr & = {\cos ^2}x.\left\{ {\frac{1}{2} - \log \,\cos \,x} \right\} + k \cr} $$

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

$$\eqalign{ & \int {\frac{{x - 1}}{{{{\left( {x + 1} \right)}^2}\sqrt {{x^3} + {x^2} + x} }}} dx \cr & = \int {\frac{{\left( {{x^2} - 1} \right)}}{{\left( {{x^2} + 2x + 1} \right).x\sqrt {x + \frac{1}{x} + 1} }}} dx \cr & = \int {\frac{{\left( {1 - \frac{1}{{{x^2}}}} \right)}}{{\left( {x + \frac{1}{x} + 2} \right)\sqrt {x + \frac{1}{x} + 1} }}dx} \cr & = \int {\frac{{2z\,dz}}{{\left( {{z^2} + 1} \right).z}}} \,\,\,\left[ {{\text{Putting }}x + \frac{1}{x} + 1 = {z^2} \Rightarrow \left( {1 - \frac{1}{{{x^2}}}} \right)dx = 2z\,dz} \right] \cr & = 2\int {\frac{{dz}}{{1 + {z^2}}}} \cr & = 2\,{\tan ^{ - 1}}z + C \cr & = 2\,{\tan ^{ - 1}}\sqrt {\frac{{{x^2} + x + 1}}{x}} + C \cr} $$

$$\eqalign{ & \int {\frac{{dx}}{{\cos \,x - \sin \,x}}} \cr & = \int {\frac{{dx}}{{\sqrt 2 \cos \left( {x + \frac{\pi }{4}} \right)}}} \cr & = \frac{1}{{\sqrt 2 }}\int {\sec \left( {x + \frac{\pi }{4}} \right)dx} \cr & = \frac{1}{{\sqrt 2 }}\log \left| {\tan \left( {\frac{\pi }{4} + \frac{x}{2} + \frac{\pi }{8}} \right)} \right| + C \cr & \left[ {\because \int {\sec \,x\,dx = \log \,\left| {\tan \left( {\frac{\pi }{4} + \frac{x}{2}} \right)} \right|} } \right] \cr & = \frac{1}{{\sqrt 2 }}\log \left| {\tan \left( {\frac{x}{2} + \frac{{3\pi }}{8}} \right)} \right| + C \cr} $$

$$\eqalign{ & f\left( x \right) = \int {\frac{{{x^2}\left( {\sqrt {1 + {x^2}} - 1} \right)}}{{\left( {1 + {x^2}} \right)\left( {1 + {x^2} - 1} \right)}}dx} \cr & = \int {\frac{{\sqrt {1 + {x^2}} - 1}}{{1 + {x^2}}}dx} \cr & = \int {\frac{{dx}}{{\sqrt {1 + {x^2}} }} - } \int {\frac{{dx}}{{1 + {x^2}}}} \cr & = \log \left( {x + \sqrt {1 + {x^2}} } \right) - {\tan ^{ - 1}}x + k \cr & \therefore f\left( 0 \right) = \log \,1 - {\tan ^{ - 1}}0 + k = k = 0 \cr & \therefore f\left( x \right) = \log \left( {x + \sqrt {1 + {x^2}} } \right) - {\tan ^{ - 1}}x \cr & \therefore f\left( 1 \right) = \log \left( {1 + \sqrt 2 } \right) - \frac{\pi }{4} \cr} $$

$$\eqalign{ & I = \int {\frac{{x{e^x}}}{{\sqrt {1 + {e^x}} }}} dx{\text{ we have}} \cr & \int {\frac{{{e^x}}}{{\sqrt {1 + {e^x}} }}} dx = 2\sqrt {1 + {e^x}} \cr} $$
Integrating $$I$$ by parts with $$x$$ as first function
$$\eqalign{ & I = x.2\sqrt {1 + {e^x}} - \int {2\sqrt {1 + {e^x}} } dx \cr & = 2x\sqrt {1 + {e^x}} - 2\int {t.\frac{{2t\,dt}}{{{t^2} - 1}}\,\,\,\,\,\,} \left( {{\text{Putting }}1 + {e^x} = {t^2}} \right) \cr & = 2x\sqrt {1 + {e^x}} - 4\int {\frac{{{t^2} - 1 + 1}}{{{t^2} - 1}}dt} \cr & = 2x\sqrt {1 + {e^x}} - 4\left[ {t + \frac{1}{2}\log \frac{{t - 1}}{{t + 1}}} \right] + c \cr & = 2\left( {x - 2} \right)\sqrt {1 + {e^x}} - 2\,\log \left( {\frac{{\sqrt {1 + {e^x}} - 1}}{{\sqrt {1 + {e^x}} + 1}}} \right) + c \cr} $$

$$\eqalign{ & \int {\frac{{dx}}{{\sin \,x\left( {3 + {{\cos }^2}x} \right)}}} \cr & = \int {\frac{{\sin \,x\,dx}}{{{{\sin }^2}\,x\left( {3 + {{\cos }^2}x} \right)}}} \cr & = \int {\frac{{\sin \,x\,dx}}{{\left( {1 - {{\cos }^2}x} \right)\left( {3 + {{\cos }^2}x} \right)}}} \cr & = \int {\frac{{dy}}{{\left( {{y^2} - 1} \right)\left( {{y^2} + 3} \right)}}\,\,\,\,\,\left( {{\text{Putting }}\cos \,x = y} \right)} \cr & = \frac{1}{4}\int {\left[ {\frac{1}{{{y^2} - 1}} - \frac{1}{{{y^2} + 3}}} \right]dy} \cr & = \frac{1}{4}\log \left| {\frac{{y - 1}}{{y + 1}}} \right| - \frac{1}{{4\sqrt 3 }}{\tan ^{ - 1}}\frac{y}{{\sqrt 3 }} + C \cr} $$

$$\eqalign{ & I = \int {{e^x}.\frac{{1 + \sin \,x}}{{2{{\cos }^2}\frac{x}{2}}}dx} \cr & \,\,\,\,\, = \int {{e^x}\left\{ {\frac{1}{2}{{\sec }^2}\frac{x}{2} + \tan \frac{x}{2}} \right\}dx} \cr & \,\,\,\,\, = {e^x}.\tan \frac{x}{2} + k \cr} $$

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

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