Definite Integration MCQ Questions & Answers in Calculus | Maths

$$\eqalign{ & I = \int\limits_{ - \pi }^\pi {\frac{{{{\cos }^2}x}}{{1 + {a^x}}}dx} .....(1) \cr & {\text{Put }}x = - y{\text{ and }}dx = - dy \cr & I = \int\limits_\pi ^{ - \pi } {\frac{{{{\cos }^2}y}}{{1 + {a^{ - y}}}}dy} = \int\limits_{ - \pi }^\pi {\frac{{{a^y}{{\cos }^2}y}}{{1 + {a^y}}}dy} \cr & I = \int\limits_{ - \pi }^\pi {\frac{{{a^x}{{\cos }^2}x}}{{1 + {a^x}}}dx} .....(2) \cr & \left[ {\because \int_a^b {f\left( y \right)dy = } \int_a^b {f\left( x \right)dx} } \right] \cr & {\text{Adding (1) and (2), we get}} \cr & 2I = \int_{ - \pi }^\pi {\frac{{\left( {1 + {a^x}} \right){{\cos }^2}x}}{{\left( {1 + {a^x}} \right)}}dx} = \int_{ - \pi }^\pi {{{\cos }^2}x\,dx} \cr & 2I = \int_0^\pi {{{\cos }^2}x\,dx} \,\,\,\,\left[ {{\text{even function}}} \right] \cr & I = 2\int_0^{\frac{\pi }{2}} {{{\cos }^2}x\,dx} .....(3) \cr & \left[ {\because \int\limits_0^{2a} {f\left( x \right)dx} = 2\int\limits_0^a {f\left( x \right)dx} \,{\text{if }}f\left( {2a - x} \right) = f\left( x \right)} \right] \cr & = 2\int\limits_0^{\frac{\pi }{2}} {{{\sin }^2}x\,dx} .....(4) \cr & {\text{Adding (3) and (4), we get}} \cr & 2I = 2\int\limits_0^{\frac{\pi }{2}} {\left( {co{s^2}x + {{\sin }^2}x} \right)dx = 2.\frac{\pi }{2} = \pi } \cr & \therefore I = \frac{\pi }{2} \cr} $$

$$\eqalign{ & {\text{Put}}\,\,{\log _e}x = z \cr & {\text{Then }}I = \int_0^{37} {\pi .\sin \,\pi z\,dz} \cr & = \left[ { - \cos \,\pi z} \right]_0^{37} \cr & = - \cos \,37\pi + \cos \,0 \cr & = 1 + 1 \cr & = 2 \cr} $$

$$\eqalign{ & \int\limits_{ - \frac{\pi }{2}}^{\frac{\pi }{2}} {\left[ {{x^2} + ln\left( {\frac{{\pi + x}}{{\pi - x}}} \right)} \right]} \cos x\,dx \cr & = \int\limits_{ - \frac{\pi }{2}}^{\frac{\pi }{2}} {{x^2}} \cos x\,dx + \int\limits_{ - \frac{\pi }{2}}^{\frac{\pi }{2}} {ln\left( {\frac{{\pi + x}}{{\pi - x}}} \right)} \cos x\,dx \cr & = 2\int\limits_0^{\frac{\pi }{2}} {{x^2}} \cos \,x\,dx + 0 \cr} $$
[as $${{x^{2\,}}\,\cos \,x}$$    is an even function and $$ln\left( {\frac{{\pi + x}}{{\pi - x}}} \right)\,\cos \,x$$     is an odd function]
$$\eqalign{ & = 2\left[ {{x^2}\sin \,x + 2x\,\cos \,x - 2\,\sin x} \right]_0^{\frac{\pi }{2}} \cr & = 2\left( {\frac{{{\pi ^2}}}{4} - 2} \right) \cr & = \frac{{{\pi ^2}}}{2} - 4 \cr} $$

$$\eqalign{ & {\text{Let }}g\left( x \right) = z \cr & {\text{Then }}I = \int_{g\left( 1 \right)}^{g\left( 2 \right)} {\frac{1}{{f\left( z \right)}}} .f'\left( z \right)dz \cr & = \left[ {\log \,f\left( z \right)} \right]_{g\left( 1 \right)}^{g\left( 2 \right)} \cr & = \log \,f\left\{ {g\left( 2 \right)} \right\} - \log \,f\left\{ {g\left( 1 \right)} \right\} \cr & = 0\,\,\,\,\,\,\,\,\left[ {\because \,g\left( 1 \right) = g\left( 2 \right)} \right] \cr} $$

$$\eqalign{ & {\text{Given }}f\left( x \right) = f\left( x \right) + f\left( {\frac{1}{x}} \right),{\text{where }}f\left( x \right) = \int_1^x {\frac{{\log \,t}}{{1 + t}}dt} \cr & \therefore F\left( e \right) = f\left( e \right) + f\left( {\frac{1}{e}} \right) \cr & \Rightarrow F\left( e \right) = \int_1^e {\frac{{\log \,t}}{{1 + t}}dt + \int_1^{\frac{1}{e}} {\frac{{\log \,t}}{{1 + t}}dt} .....({\text{A}})\,} \cr & {\text{Now for solving ,}}\,\,\,I = \int_1^{\frac{1}{e}} {\frac{{\log \,t}}{{1 + t}}dt} \cr & \therefore {\text{Put }}\frac{1}{t} = z\, \Rightarrow - \frac{1}{{{t^2}}}dt = dz\, \Rightarrow dt = - \frac{{dz}}{{{z^2}}} \cr & {\text{and limit for }}t = 1\, \Rightarrow z = 1\,{\text{and for }}t = \frac{1}{e} \Rightarrow z = e \cr & \therefore I = \int_1^e {\frac{{\log \left( {\frac{1}{z}} \right)}}{{1 + \frac{1}{z}}}\left( { - \frac{{dz}}{{{z^2}}}} \right)} \cr & = \int_1^e {\frac{{\left( {\log \,1 - \log \,z} \right).z}}{{z + 1}}} \left( { - \frac{{dz}}{{{z^2}}}} \right) \cr & = \int_1^e { - \frac{{\log \,z}}{{\left( {z + 1} \right)}}\left( { - \frac{{dz}}{z}} \right)} \cr & = \int_1^e {\frac{{\log \,z}}{{z\left( {z + 1} \right)}}dz} \,\,\,\left[ {\therefore \log \,1 = 0} \right] \cr & \therefore I = \int_1^e {\frac{{\log \,t}}{{t\left( {t + 1} \right)}}dt} \cr & \left[ {{\text{By property }}\int_a^b {f\left( t \right)dt} = \int_a^b {f\left( x \right)dx} } \right] \cr & {\text{Equation (A) becomes}} \cr & F\left( e \right) = \int_1^e {\frac{{\log \,t}}{{1 + t}}dt + } \int_1^e {\frac{{\log \,t}}{{t\left( {1 + t} \right)}}dt} \cr & = \int_1^e {\frac{{t.\log \,t + \log \,t}}{{t\left( {1 + t} \right)}}dt} \cr & = \int_1^e {\frac{{\left( {\log \,t} \right)\left( {t + 1} \right)}}{{t\left( {1 + t} \right)}}dt} \cr & \Rightarrow F\left( e \right) = \int_1^e {\frac{{\log \,t}}{t}dt} \cr & {\text{Let }}\log \,t = x \cr & \therefore \frac{1}{t}dt = dx \cr & \left[ {{\text{for time }}t = 1,\,x = 0{\text{ and }}t = e,x = \log \,e = 1} \right] \cr & \therefore F\left( e \right) = \int_0^1 {x\,dx} \cr & F\left( e \right) = \left[ {\frac{{{x^2}}}{2}} \right]_0^1\, \Rightarrow F\left( e \right) = \frac{1}{2} \cr} $$

$$\eqalign{ & p'\left( x \right) = p'\left( {1 - x} \right)\,\, \Rightarrow p\left( x \right) = - p\left( {1 - x} \right) + c \cr & {\text{At }}x = 0 \cr & p\left( 0 \right) = - p\left( 1 \right) + c\,\, \Rightarrow 42 = c \cr & {\text{Now }}p\left( x \right) = - p\left( {1 - x} \right) + 42 \cr & \Rightarrow p\left( x \right) + p\left( {1 - x} \right) = 42 \cr & \Rightarrow I = \int\limits_0^1 {p\left( x \right)dx} .....{\text{(i)}} \cr & \Rightarrow I = \int\limits_0^1 {p\left( {1 - x} \right)dx} .....{\text{(ii)}} \cr} $$
on adding (i) and (ii), we get
$$2I = \int\limits_0^1 {\left( {42} \right)dx} \,\, \Rightarrow I = 21$$

$$\eqalign{ & I = \int\limits_0^\pi {x\,f\left( {\sin \,x} \right)dx} \cr & \Rightarrow I = \int\limits_0^\pi {\left( {\pi - x} \right)f\left( {\sin \,x} \right)dx} \cr & \Rightarrow I = \pi \int\limits_0^\pi {f\left( {\sin \,x} \right)dx - I} \cr & \Rightarrow 2I = \pi \int\limits_0^\pi {f\left( {\sin \,x} \right)dx} \cr & \Rightarrow I = \frac{\pi }{2}\int\limits_0^\pi {f\left( {\sin \,x} \right)dx} \cr & \Rightarrow I = \pi \int\limits_0^{\frac{\pi }{2}} {f\left( {\sin \,x} \right)dx} \cr & \Rightarrow I = \pi \int\limits_0^{\frac{\pi }{2}} {f\left( {\cos \,x} \right)dx} \cr} $$

$$\eqalign{ & {\text{Let }}I = \int\limits_{\frac{\pi }{4}}^{\frac{\pi }{2}} {{{\left( {2\,{\text{cosec}}\,x} \right)}^{17}}dx} \cr & = \int\limits_{\frac{\pi }{4}}^{\frac{\pi }{2}} {{{\left( {\,{\text{cosec}}\,x + \cot \,x + {\text{cosec}}\,x - \cot \,x} \right)}^{16}}{\text{2}}\,{\text{cosec}}\,x\,dx} \cr & I = 2\int\limits_{\frac{\pi }{4}}^{\frac{\pi }{2}} {{{\left( {{\text{cosec}}\,x + \cot \,x + \frac{1}{{{\text{cosec}}\,x + \cot \,x}}} \right)}^{16}}.\,{\text{cosec}}\,x\,dx} \cr & {\text{Let cosec}}\,x + \cot \,x = {e^u} \cr & \Rightarrow \left( { - \,{\text{cosec}}\,x\,\cot \,x - {\text{cosec}}{\,^2}x} \right)dx = {e^u}\,du \cr & \Rightarrow - \,{\text{cosec}}\,x\,dx = du \cr & {\text{Also at }}x = \frac{\pi }{4}{\text{,}}\,u = ln\left( {\sqrt 2 + 1} \right) \cr & {\text{at }}x = \frac{\pi }{2},\,u = ln\,1 = 0 \cr & \therefore I = - 2\,\int\limits_{ln\left( {\sqrt 2 + 1} \right)}^0 {{{\left( {{e^u} + {e^{ - u}}} \right)}^{16}}du} \cr & = 2\int\limits_0^{ln\left( {\sqrt 2 + 1} \right)} {{{\left( {{e^u} + {e^{ - u}}} \right)}^{16}}} du \cr} $$

$${\text{Here }}f\left( x \right) = \int\limits_1^x {\sqrt {2 - {t^2}} dt \Rightarrow f'\left( x \right) = \sqrt {2 - {x^2}} } $$
Now the given equation $${x^2} - f'\left( x \right) = 0$$    becomes
$$\eqalign{ & {x^2} - \sqrt {2 - {x^2}} = 0 \cr & \Rightarrow {x^2} = \sqrt {2 - {x^2}} \cr & \Rightarrow x = \pm 1 \cr} $$

$$\eqalign{ & {\text{Let }}f\left( x \right) = \frac{{\left| x \right|dx}}{{8\,{{\cos }^2}2x + 1}} \cr & {\text{then }}f\left( { - x} \right) = \frac{{\left| { - x} \right|}}{{8\,{{\cos }^2}2\left( { - x} \right) + 1}} = \frac{{\left| { - x} \right|}}{{8\,{{\cos }^2}2x + 1}} = f\left( x \right) \cr & \therefore \,f\left( x \right){\text{ is even function}} \cr & \therefore \,I = \int\limits_{\frac{{ - \pi }}{2}}^{\frac{\pi }{2}} {\frac{{\left| x \right|dx}}{{8\,{{\cos }^2}2x + 1}}} \cr & \Rightarrow I = 2\int\limits_0^{\frac{\pi }{2}} {\frac{{\left| x \right|dx}}{{8\,{{\cos }^2}2x + 1}}} \cr & \Rightarrow I = 2\int\limits_0^{\frac{\pi }{2}} {\frac{{x\,dx}}{{8\,{{\cos }^2}2x + 1}}} \cr & \Rightarrow I = 2{I_1} \cr & {\text{Now,}} \cr & \therefore {\text{ }}{I_1} = \int\limits_0^{\frac{\pi }{2}} {\frac{{\left( {\frac{\pi }{2} - x} \right)dx}}{{8\,{{\cos }^2}2\left( {\frac{\pi }{2} - x} \right) + 1}}} \cr & \Rightarrow {I_1} = \int\limits_0^{\frac{\pi }{2}} {\frac{{\frac{\pi }{2} - x}}{{8\,{{\cos }^2}2x + 1}}dx} \cr & \Rightarrow {I_1} = \frac{\pi }{2}\int\limits_0^{\frac{\pi }{2}} {\frac{{dx}}{{8\,{{\cos }^2}2x + 1}}} - {I_1} \cr & \Rightarrow 2{I_1} = \frac{\pi }{2}.2\int\limits_0^{\frac{\pi }{4}} {\frac{{dx}}{{8\,{{\cos }^2}2x + 1}}} \cr & \Rightarrow 2{I_1} = \pi \int\limits_0^{\frac{\pi }{4}} {\frac{{{{\sec }^2}2x}}{{9 + {{\tan }^2}2x}}} dx\,; \cr & {\text{Put }}\tan \,2x = t\,\, \Rightarrow 2{\sec ^2}2x\,dx = dt \cr & \Rightarrow 2{I_1} = \frac{\pi }{2}\int\limits_0^\infty {\frac{{dt}}{{9 + {t^2}}}} \cr & \Rightarrow 2{I_1} = \frac{\pi }{2}.\frac{1}{3}\left[ {{{\tan }^{ - 1}}\frac{t}{3}} \right]_0^\infty \cr & \Rightarrow 2{I_1} = \frac{\pi }{2}.\frac{1}{3}.\frac{\pi }{2} \cr & \therefore \,{I_1} = \frac{{{\pi ^2}}}{{24}} \Rightarrow I = 2{I_1} = \frac{{{\pi ^2}}}{{12}} \cr} $$