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This is one of our many quizzes on Integration by substitution. Click on this to access more such competitive quizzes on Integral Calculus.

Integration by substitution Quiz 2

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 Write the value of integral$\int{\sqrt{1+\cos 11p}}.\cos \dfrac{11p}{2}dp$.

 Use the identity $1+\cos \theta =2{{\cos }^{2}}\dfrac{\theta }{2}$ and then solve further.

  $\begin{align}  & \text{ Write the given integral as I.} \\  & I=\int{\sqrt{1+\cos 11p}.\cos \dfrac{11}{2}pdp} \\  & \Rightarrow I=\int{\sqrt{2{{\cos }^{2}}\left( \dfrac{11p}{2} \right)}.\cos \dfrac{11}{2}pdp}\text{        }\left[ \because 1+\cos \theta =2{{\cos }^{2}}\dfrac{\theta }{2} \right] \\  & =\int{\sqrt{2}.{{\cos }^{2}}\dfrac{11p}{2}dp} \\  & =\sqrt{2}\int{\dfrac{1+\cos 11p}{2}}dp\text{        }\left[ \because {{\cos }^{2}}\theta =\dfrac{1+\cos 2\theta }{2} \right] \\  & \Rightarrow \dfrac{\sqrt{2}}{2}\left[ p+\dfrac{\sin 11p}{11} \right]+c \\  & \Rightarrow I=\dfrac{\sqrt{2}}{2}p+\dfrac{\sqrt{2}}{22}\sin 11p+c \\ \end{align}$

 Solve the integral $\int{\sqrt{19t+5}}dt$ by using the substitution method.

 Substitute  $19t+5=u$ and use a substitution method to solve the integral.

  $\begin{align}  & \text{Let }I=\int{\sqrt{19t+5}}dt .....\left( i \right) \\  & Put 19t+5=u \\  & \Rightarrow 19=\dfrac{du}{dt}\Rightarrow 19dt=du \\  & \Rightarrow dt=\dfrac{du}{19} .....\left( ii \right) \\  & \text{Use equation }\left( ii \right)\text{ in equation }\left( i \right) \\  & I=\dfrac{1}{19}\int{\sqrt{u}}du \\  & I=\dfrac{1}{19}.\dfrac{{{u}^{\dfrac{3}{2}}}}{\dfrac{3}{2}}+c=\dfrac{2}{57}{{u}^{\dfrac{3}{2}}}+c \\  & Substitute u=19t+5 \\  & \therefore I=\dfrac{2}{57}{{\left( 19t+5 \right)}^{\dfrac{3}{2}}}+c \\ \end{align}$

 Write the value of integral $\int_{0}^{1}{\dfrac{3{{x}^{2}}+2x+1}{{{\left( {{x}^{3}}+{{x}^{2}}+x+1 \right)}^{3}}}dx}$

 Substitute ${{x}^{3}}+{{x}^{2}}+x+1=t$ and use the substitution method to solve the integral. Then apply the limits.

$\begin{align}  & \text{Let }I=\int_{0}^{1}{\dfrac{3{{x}^{2}}+2x+1}{{{\left( {{x}^{3}}+{{x}^{2}}+x+1 \right)}^{3}}}dx} .....\left( i \right) \\  & \text{Put }{{x}^{3}}+{{x}^{2}}+x+1=t \\  & \Rightarrow \left( 3{{x}^{2}}+2x+1 \right)dx=dt .....\left( ii \right) \\  & \text{Use equation }\left( ii \right)\text{ in equation }\left( i \right) \\  & I=\int_{0}^{1}{\dfrac{dt}{{{t}^{3}}}} \\  & =\left[ -\dfrac{1}{2{{t}^{2}}} \right]_{0}^{1}=\left[ \dfrac{-1}{2{{\left( {{x}^{3}}+{{x}^{2}}+x+1 \right)}^{2}}} \right]_{0}^{1}=\dfrac{15}{32} \\ \end{align}$

 Evaluate $\int_{0}^{\dfrac{1}{12}}{\dfrac{{{\sin }^{-1}}\left( 21x \right)}{\sqrt{1-441{{x}^{2}}}}}dx$

 Put  ${{\sin }^{-1}}\left( 21x \right)=t$ and use a substitution method to solve the integral. Then apply the limits.

 $\begin{align}  & \text{Consider the integral as I.} \\  & I=\int_{0}^{\dfrac{1}{21}}{\dfrac{{{\sin }^{-1}}\left( 21x \right)}{\sqrt{1-441{{x}^{2}}}}dx} .....\left( i \right) \\  & \text{Put }{{\sin }^{-1}}\left( 21x \right)=t \\  & \Rightarrow \dfrac{1}{\sqrt{1-441{{x}^{2}}}}.21=\dfrac{dt}{dx} \\  & \Rightarrow \dfrac{1}{\sqrt{1-441{{x}^{2}}}}dx=\dfrac{dt}{21} .....\left( ii \right) \\  & \text{Put value of }\dfrac{dx}{\sqrt{1-441{{x}^{2}}}} \text{ from equation }\left( ii \right)\text{ in equation }\left( i \right) \\  & I=\dfrac{1}{21}\int_{0}^{\dfrac{1}{21}}{t}dt \\  & =\dfrac{1}{21}.\dfrac{{{t}^{2}}}{2}+c={{\left[ \dfrac{{{t}^{2}}}{42} \right]}^{\dfrac{1}{21}}} \\  & \Rightarrow I\left[ {{\dfrac{\left[ {{\sin }^{-1}}(21x) \right]}{42}}^{2}} \right]_{0}^{\dfrac{1}{21}}\text{            }\left[ \because t={{\sin }^{-1}}\left( 21x \right) \right] \\  & \text{Further solve} \\  & I=\dfrac{{{\left[ {{\sin }^{-1}}\left( 1 \right) \right]}^{2}}}{42}-{{\dfrac{\left[ {{\sin }^{-1}}\left( 0 \right) \right]}{42}}^{2}} \\  & I=\dfrac{{{\left( \dfrac{\pi }{2} \right)}^{2}}}{42}=\dfrac{{{\pi }^{2}}}{168} \\ \end{align}$

 Simply and write the value of $\int{\dfrac{{{3}^{z}}\log 3+{{z}^{z}}(1+\log z)+3{{z}^{2}}}{{{3}^{z}}+{{z}^{3}}+{{z}^{z}}}}dz$.

 Differentiate  ${{z}^{z}}$ using logarithms. Then put ${{3}^{z}}+{{z}^{3}}+{{z}^{z}}=t$ and use a substitution method to solve the integral.

 $\begin{align}  & \text{Let }I=\int{\dfrac{{{3}^{z}}\log 3+{{z}^{z}}\left( 1+\operatorname{logz} \right)+3{{z}^{2}}}{{{3}^{z}}+{{z}^{3}}+{{z}^{z}}}}dz .....\left( i \right) \\  & \text{Put }{{3}^{z}}+{{z}^{3}}+{{z}^{z}}=t \\  & \text{Differentiate both sides write ‘z’} .\\  & {{3}^{z}}\log 3+3{{z}^{2}}+{{z}^{z}}\left( 1+\log z \right)=\dfrac{dt}{dz} \\  & Let u={{z}^{z}} \\  & \Rightarrow \operatorname{logu}=log{{z}^{z}} \\  & \Rightarrow \log u=z\log z \\  & \text{Differentiate write }\dfrac{1}{u}\dfrac{du}{dz}=z\times \dfrac{1}{z}+\log z\times 1 \\  & \Rightarrow \dfrac{du}{dz}=u\left( 1+\log z \right)={{z}^{z}}\left( 1+\log z \right) \\  & \Rightarrow \left[ {{3}^{z}}\log 3+3{{z}^{2}}+{{z}^{z}}\left( 1+\log z \right) \right]dz=dt .....\left( ii \right) \\  & \text{Use equation }\left( ii \right)\text{ in equation }\left( i \right) \\  & I=\int{\dfrac{dt}{t}} \\  & =\log t+c \\  & \text{Substitute }t={{3}^{z}}+{{z}^{3}}+{{z}^{z}} \\  & \therefore I=\log \left| {{3}^{z}}+{{z}^{3}}+{{z}^{z}} \right|+c \\ \end{align}$