ONE MORE STEP TOWARD BETTER TOMORROW : CHAPTER 13- Worked Out Examples

Monday, 4 August 2014

Example: 1

Evaluate the following integrals

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:

	$I = \int {{{\mathop {\tan }\limits_{{\rm{Ist}}} }^{ - 1}}x \cdot \mathop x\limits_{{\rm{IInd}}} \,\,dx}$
	$\,\,\,\, = \dfrac{{{x^2}}}{2}{\tan ^{ - 1}}x - \dfrac{1}{2}\int {\dfrac{1}{{1 + {x^2}}} \cdot {x^2}\,\,dx}$
	$\,\,\,\, = \dfrac{{{x^2}}}{2}{\tan ^{ - 1}}x - \dfrac{1}{2}\int {\dfrac{{1 + {x^2} - 1}}{{1 + {x^2}}}\,\,dx}$
	$\,\,\,\, = \dfrac{{{x^2}}}{2}{\tan ^{ - 1}}x - \dfrac{1}{2}\left\{ {\int {dx - \int {\dfrac{1}{{1 + {x^2}}}} \,\,dx} } \right\}$
	$\,\,\,\, = \dfrac{{{x^2}}}{2}{\tan ^{ - 1}}x - \dfrac{x}{2} + \dfrac{1}{2}{\tan ^{ - 1}}x + C$

Monday, 4 August 2014