Examples On Integration By Parts Set-1
Example – 41
Evaluate the following integrals
| (a) \(\int {x\sin x\,\,dx} \) | (b) \(\int {{{(\ln x)}^2}\,\,dx} \) |
| (c) \(\int {{{\sin }^{ - 1}}x\,\,dx} \) | (d) \(\int {x \;{{\tan }^{ - 1}}x\,dx} \) |
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:
| (a) \(\int {\theta {{\sec }^2}\theta \,\,d\theta } \) | (b) \(\int {{e^x}\sin x\,\,dx} \) |
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:
\[\begin{align}&\qquad\qquad\;\; I = \int {\mathop {\sin }\limits_{{\rm{Ist}}} x \cdot {{\mathop e\limits_{{\rm{IInd}}} }^x}\,\,dx}\\&\qquad\qquad\;\;\;\; = \sin x \cdot {e^x} - \int {\mathop {\cos }\limits_{{\rm{Ist}}} x \cdot {{\mathop e\limits_{{\rm{IInd}}} }^x}dx} \\&\qquad\qquad\;\;\;\; = {e^x}\sin x - \left\{ {\cos x \cdot {e^x} - \int {\left( { - \sin x} \right){e^x}\,\,dx} } \right\}\\&\qquad\qquad\;\;\;\; = {e^x}\sin x - {e^x}\cos x - \int {{e^x}} \sin x\,\,dx\\&\;\;\;\;\qquad\qquad = {e^x}(\sin x - \cos x) - I\\ &\Rightarrow\quad\quad\;\; 2I = {e^x}(\sin x - \cos x)\\ & \Rightarrow \quad\quad\quad I = \frac{1}{2}{e^x}(\sin x - \cos x) + C\end{align}\]