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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
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