tgt

## Monday, 4 August 2014

### CHAPTER 13- Worked Out Examples

 Example: 1
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: 1-(a)
Using the $ILATE$ rule, we let $f(x) = x$ be our first function:
 $I = \int {\mathop x\limits_{{\rm{Ist}}} \mathop {\sin }\limits_{{\rm{IInd}}} x\,\,dx}$ $\,\,\,\, = - x\cos x - \int {1 \cdot ( - \cos x)\,\,dx}$ $\,\,\,\, = - x\cos x + \sin x + C$
See what happens if you choose $\sin x$ as the first function.
 Solution: 1-(b)
We choose unity as the second function and apply integration by parts:
 $I = \int {\mathop {{{(\ln x)}^2}}\limits_{{\rm{Ist}}} \cdot \mathop 1\limits_{{\rm{IInd}}} \,\,dx}$ $\,\,\,\, = x{(\ln x)^2} - \int {\left( {\dfrac{{2\ln x}}{x}} \right) \cdot x\,\,dx}$ $= x{(\ln x)^2} - 2\int {\underbrace{\ln x}_{\rm{Comment:}}dx}$
Comment: Again apply integration by parts, taking unity as the second function
 $\,\,\,\, = x{(\ln x)^2} - 2\left\{ {x\ln x - \int {\dfrac{1}{x} \cdot x\,\,\,dx} } \right\}$ $\,\,\,\, = x{(\ln x)^2} - 2x\ln x + 2x + C$
 Solution: 1-(c)
Here again, we choose unity as the second function:
 $I = \int {{{\mathop {\sin }\limits_{{\rm{Ist}}} }^{ - 1}}x \cdot \mathop 1\limits_{{\rm{IInd}}} \,\,dx}$ $= x{\sin ^{ - 1}}x - \int {\dfrac{1}{{\sqrt {\underbrace{1 - {x^2}}_{\rm{{Comment}}}} }}} .xdx$ Comment: Substitute $1 - {x^2} = t \,\,\, \Rightarrow\,\,\,\, x\,\,dx = dt/2$ $\,\,\,\, = x{\sin ^{ - 1}}x + \dfrac{1}{2}\int {\dfrac{{dt}}{{\sqrt t }}}$ $\,\,\,\, = x{\sin ^{ - 1}}x + \sqrt t + C$ $\,\,\,\, = x{\sin ^{ - 1}}x + \sqrt {1 - {x^2}} + C$
 Solution: 1-(d)
Using the $ILATE$ rule, we choose ${\tan ^{ - 1}}x$ as the first function:
 $I = \int {{{\mathop {\tan }\limits_{{\rm{Ist}}} }^{ - 1}}x \cdot \mathop x\limits_{{\rm{IInd}}} \,\,dx}$ $\,\,\,\, = \dfrac{{{x^2}}}{2}{\tan ^{ - 1}}x - \dfrac{1}{2}\int {\dfrac{1}{{1 + {x^2}}} \cdot {x^2}\,\,dx}$ $\,\,\,\, = \dfrac{{{x^2}}}{2}{\tan ^{ - 1}}x - \dfrac{1}{2}\int {\dfrac{{1 + {x^2} - 1}}{{1 + {x^2}}}\,\,dx}$ $\,\,\,\, = \dfrac{{{x^2}}}{2}{\tan ^{ - 1}}x - \dfrac{1}{2}\left\{ {\int {dx - \int {\dfrac{1}{{1 + {x^2}}}} \,\,dx} } \right\}$ $\,\,\,\, = \dfrac{{{x^2}}}{2}{\tan ^{ - 1}}x - \dfrac{x}{2} + \dfrac{1}{2}{\tan ^{ - 1}}x + C$