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This is one of our many quizzes on Techniques of integration. Click on this to access more such competitive quizzes on Calculus.

Techniques of integration Quiz 2

mathematics

 what will be the form of integral after the second term substitution in $\int{\left( \dfrac{{{e}^{x}}}{{{e}^{x}}+{{e}^{-x}}} \right)dx}$?

 Integrate the terms by simplifying them and then using the substitution method and solve the integral.

  $I=\int{\left( \dfrac{{{e}^{x}}}{{{e}^{x}}+{{e}^{-x}}} \right)dx} \\ \Rightarrow \dfrac{1}{2}\int{\left( \int{\left( \dfrac{{{e}^{x}}+{{e}^{-x}}}{{{e}^{x}}+{{e}^{-x}}}+\dfrac{{{e}^{x}}-{{e}^{x}}}{{{e}^{x}}+{{e}^{-x}}} \right)dx} \right)} \\ \text{Let }t={{e}^{x}}+{{e}^{-x}},dt=\left( {{e}^{x}}-{{e}^{-x}} \right)dx \\ \Rightarrow \int{\dfrac{dx}{2}+\dfrac{1}{2}\int{\dfrac{dt}{t}}} \\ \dfrac{1}{2}\int{\dfrac{dt}{t}}=\int{\dfrac{dv}{2v}}$

 When using partial fraction to integrate $\int\limits_{0}^{1}{\dfrac{x-4}{{{x}^{2}}-5x+6}}\ dx$, what will be the integration of the term with  $-1$ as coefficient in the split terms are?

 Since the coefficient of -1 is asked first simplify the integral so as to obtain an expression consisting a term with -1 as coefficient and then evaluate the integral.

  $I=\int\limits_{0}^{1}{\dfrac{x-4}{{{x}^{2}}-5x+6}}\ dx \\ =\int\limits_{0}^{1}{\left( \dfrac{x-3}{\left( x-3 \right)\left( x-2 \right)}-\dfrac{1}{\left( x-3 \right)\left( x-2 \right)}\  \right)}dx \\ \text{Since coefficient of -1 terms are asked: } \\ =\int\limits_{0}^{1}{\dfrac{dx}{\left( x-3 \right)\left( x-2 \right)}} \\ \Rightarrow \int\limits_{0}^{1}{\dfrac{dx}{\left( x-3 \right)}-\int\limits_{0}^{1}{\dfrac{dx}{\left( x-2 \right)}}} \\ \Rightarrow \ln {{\left| x-3 \right|}_{0}}^{1}-\ln {{\left| x-2 \right|}_{0}}^{1} \\ \Rightarrow 2\ln 2-\ln 3$

 Find the sum of constant coefficients of the terms of the integration of $\left( {{x}^{2}}+1 \right){{e}^{2x}}$ by using by parts.

 First evaluate the integral by simplifying them and using integration by parts to solve the integral then after that take the sum of the constant terms present in the integral.

  $ I=\int{\left( {{x}^{2}}+1 \right){{e}^{2x}}} \\ \text{Applying integration by parts to the above equation:} \\ I =\dfrac{\left( {{x}^{2}}+1 \right){{e}^{2x}}}{2}-\int{\dfrac{2x}{2}\times }{{e}^{2x}}dx \\ \Rightarrow \dfrac{\left( {{x}^{2}}+1 \right){{e}^{2x}}}{2}-\left[ \dfrac{x{{e}^{2x}}}{2}-\int{\dfrac{{{e}^{2x}}}{2}dx} \right] \\ \Rightarrow \dfrac{\left( {{x}^{2}}+1 \right){{e}^{2x}}}{2}-\left[ \dfrac{x{{e}^{2x}}}{2}-\dfrac{{{e}^{2x}}}{4} \right]+c \\ \Rightarrow {{e}^{2x}}\left[ \dfrac{{{x}^{2}}}{2}+\dfrac{1}{2}-\dfrac{x}{2}+\dfrac{1}{4} \right] \\ \text{Sum of constants }= \dfrac{1}{2}+\dfrac{1}{4}=\dfrac{3}{4}$

 What is the coefficient of $\pi $ in the integral  $\int\limits_{\frac{3}{2}}^{3}{\sqrt{9-{{x}^{2}}}}dx$ using trigonometric substitution?

 We can solve the integral by using the substitution method $\text{ }x=3\sin (u) ,dx=3\cos (u) $ and then evaluating it to see the coefficient of $\pi $.

  $I=\int\limits_{\frac{3}{2}}^{3}{\sqrt{9-{{x}^{2}}}}dx \\ \text{Let }x=3\sin(u) ,dx=3\cos(u) \\ =\int _{\frac{\pi }{6}}^{\frac{\pi }{2}}9\cos ^2\left(u\right)du \\ 9\int _{\frac{\pi }{6}}^{\frac{\pi }{2}}\dfrac{1+\cos \left(2u\right)}{2}du \\ 9 \dfrac{1}{2}\cdot \int _{\frac{\pi }{6}}^{\frac{\pi }{2}}1+\cos \left(2u\right)du \\ 9\dfrac{1}{2}\left(\int _{\frac{\pi }{6}}^{\frac{\pi }{2}}1du+\int _{\frac{\pi }{6}}^{\frac{\pi }{2}}\cos \left(2u\right)du\right) \\ 9\cdot \dfrac{1}{2}\left(\dfrac{\pi }{3}-\dfrac{\sqrt{3}}{4}\right) \\ \dfrac{12\pi}{8}-\dfrac{9\sqrt{3}}{8} \\ \text{The coefficient of }\pi \text{ is } \dfrac{3}{2}$

 Determine using long division method$\int{\dfrac{4{{x}^{2}}+x+9}{4{{x}^{2}}-9}}dx$.

 Use the long division method to simplify the terms and then use the basic algebra to get a simple equation and integrate further.

  $ I=\int{\dfrac{4{{x}^{2}}+x+9}{4{{x}^{2}}-9}}dx \\ \text{By long division method } \\ I = 4{{x}^{2}}+x+9=\left( 4{{x}^{2}}-9 \right)+\left( x+18 \right) \\ \Rightarrow \int{\dfrac{4{{x}^{2}}-9}{4{{x}^{2}}-9}dx+}\int{\dfrac{x+18}{4{{x}^{2}}-9}dx} \\ \Rightarrow \int{dx}+\int{\dfrac{x}{4{{x}^{2}}-9}dx}+\int{\dfrac{18}{4{{x}^{2}}-9}dx} \\ \Rightarrow x+\dfrac{1}{8}\ln \left| 4{{x}^{2}}-9 \right|+3\left[ \dfrac{1}{2x-3}-\dfrac{1}{2x+3} \right]dx \\ \Rightarrow x+\dfrac{1}{8}\ln \left| 4{{x}^{2}}-9 \right|+\dfrac{3}{2}\left[ \ln \left| 2x-3 \right|-\ln \left| 2x+3 \right| \right]+c \\ \Rightarrow x+\dfrac{13}{8}\ln \left| 2x-3 \right|-\dfrac{11}{8}\ln \left| 2x+3 \right|+c$