Examples On Integration By Parts Set-1

integration by parts examples

Example – 41

Evaluate the following integrals

Solution: (a) Using the ILATE rule, we let \(f(x) = x\) be our first function:

\[\begin{align} &\; I = \int {\mathop x\limits_{{\rm{Ist}}} \mathop {\sin }\limits_{{\rm{IInd}}} x\,\,dx}\\&\quad  = - x\cos x - \int {1 \cdot ( - \cos x)\,\,dx} \\&\quad = - x\cos x + \sin x + C\end{align}\]

See what happens if you choose \(\sin x\) as the first function.

(b) We choose unity as the second function and apply integration by parts:

\[\begin{align}
 & I=\int{\underset{\text{Ist}}{\mathop{{{(\ln x)}^{2}}}}\,\cdot \underset{\text{IInd}}{\mathop{1}}\,\,\,dx} \\ 
&\;\; =x{{(\ln x)}^{2}}-\int{\left( \frac{2\ln x}{x} \right)\cdot xdx} \\
& \;\; =x{{(\ln x)}^{2}}-2\int{\ln xdx} \\ 
&\qquad\qquad  \underset{\begin{matrix}
  \text{Again apply integration by}  \\
  \text{parts, taking unity as the second}  \\
  \text{function}  \\
\end{matrix}}{\mathop{\uparrow }}\, \\ 
&\;\;  =x{{(\ln x)}^{2}}-2\left\{ x\ln x-\int{\frac{1}{x}\cdot xdx} \right\} \\ 
& \;\; =x{{(\ln x)}^{2}}-2x\ln x+2x+C \\ 
\end{align}\]

(c) Here again, we choose unity as the second function:

\[\begin{align}
 & I=\int{{{\underset{\text{Ist}}{\mathop{\sin }}\,}^{-1}}x\cdot \underset{\text{IInd}}{\mathop{1}}\,\,\,dx} \\ 
& \,\,\,\,=x{{\sin }^{-1}}x-\int{\frac{1}{\sqrt{1-{{x}^{2}}}}\cdot x\,\,dx} \\ 
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\underset{\begin{smallmatrix}  \qquad\qquad
\text{Substitute }1-{{x}^{\text{2}}}=t \\
\Rightarrow -x\,dx=dt/2 
\end{smallmatrix}}{\mathop{\nwarrow }}\, \\ 
& \,\,\,\,=x{{\sin }^{-1}}x+\frac{1}{2}\int{\frac{dt}{\sqrt{t}}} \\ 
& \,\,\,\,=x{{\sin }^{-1}}x+\sqrt{t}+C \\ 
& \,\,\,\,=x{{\sin }^{-1}}x+\sqrt{1-{{x}^{2}}}+C \\ 
\end{align}\]

(d) Using the ILATE rule, we choose \({{\tan }^{-1}}x\) as the first function:

\[\begin{align}& I = \int {{{\mathop {\tan }\limits_{{\rm{Ist}}} }^{ - 1}}x \cdot \mathop x\limits_{{\rm{IInd}}} \,\,dx}\\ &\; = \frac{{{x^2}}}{2}{\tan ^{ - 1}}x - \frac{1}{2}\int {\frac{1}{{1 + {x^2}}} \cdot {x^2}\,\,dx} \\& \;= \frac{{{x^2}}}{2}{\tan ^{ - 1}}x - \frac{1}{2}\int {\frac{{1 + {x^2} - 1}}{{1 + {x^2}}}\,\,dx} \\& \; = \frac{{{x^2}}}{2}{\tan ^{ - 1}}x - \frac{1}{2}\left\{ {\int {dx - \int {\frac{1}{{1 + {x^2}}}} \,\,dx} } \right\}\\&\; = \frac{{{x^2}}}{2}{\tan ^{ - 1}}x - \frac{x}{2} + \frac{1}{2}{\tan ^{ - 1}}x + C\end{align}\]
 

Example - 42

Evaluate the following integrals:

Solution: (a) We use the ILATE rule and let \(f(\theta ) = \theta \) be our first function:

\[\begin{align}& I = \int {\mathop \theta \limits_{{\rm{Ist}}} \cdot {{\mathop {\sec }\limits_{{\rm{IInd}}} }^2}\theta \,\,d\theta }\\ &\; = \theta \tan \theta - \int {1 \cdot \tan \theta \,\,d\theta } \\&\; = \theta \tan \theta + \ln \left| {\cos \theta } \right| + C\end{align}\]

(b) We again use the ILATE rule and let \(\sin x\) be our first function:

\[\begin{align}&\qquad\qquad\;\; I = \int {\mathop {\sin }\limits_{{\rm{Ist}}} x \cdot {{\mathop e\limits_{{\rm{IInd}}} }^x}\,\,dx}\\&\qquad\qquad\;\;\;\;  = \sin x \cdot {e^x} - \int {\mathop {\cos }\limits_{{\rm{Ist}}} x \cdot {{\mathop e\limits_{{\rm{IInd}}} }^x}dx} \\&\qquad\qquad\;\;\;\; = {e^x}\sin x - \left\{ {\cos x \cdot {e^x} - \int {\left( { - \sin x} \right){e^x}\,\,dx} } \right\}\\&\qquad\qquad\;\;\;\; = {e^x}\sin x - {e^x}\cos x - \int {{e^x}} \sin x\,\,dx\\&\;\;\;\;\qquad\qquad = {e^x}(\sin x - \cos x) - I\\ &\Rightarrow\quad\quad\;\; 2I = {e^x}(\sin x - \cos x)\\ & \Rightarrow \quad\quad\quad I = \frac{1}{2}{e^x}(\sin x - \cos x) + C\end{align}\]