Solved IGNOU Assignments and Question papers: June 2019

As per given condition

$\int\limits_0^\Pi {\frac{{x\sin x}}{{1 + {{\cos }^2}x}}} = \int\limits_0^\Pi {\frac{{\left( {\Pi - x} \right)\sin \left( {\Pi - x} \right)}}{{1 + {{\cos }^2}\left( {\Pi - x} \right)}}}$
Now we can proceed

$\Rightarrow \int\limits_0^\Pi {\frac{{x\sin x}}{{1 + {{\cos }^2}x}}} = \int\limits_0^\Pi {\frac{{\left( {\Pi - x} \right)\sin x}}{{1 + {{\cos }^2}x}}}$

$\Rightarrow \int\limits_0^\Pi {\frac{{x\sin x}}{{1 + {{\cos }^2}x}}} = \int\limits_0^\Pi {\frac{{\left( {\Pi - x} \right)\sin x}}{{1 + {{\cos }^2}x}}}$

$\Rightarrow \int\limits_0^\Pi {\frac{{x\sin x}}{{1 + {{\cos }^2}x}}} = \Pi \int\limits_0^\Pi {\frac{{\sin x}}{{1 + {{\cos }^2}x}}} - \int\limits_0^\Pi {\frac{{x\sin x}}{{1 + {{\cos }^2}x}}}$

$\Rightarrow 2\int\limits_0^\Pi {\frac{{x\sin x}}{{1 + {{\cos }^2}x}}} = \Pi \int\limits_0^\Pi {\frac{{\sin x}}{{1 + {{\cos }^2}x}}}$

$\Rightarrow \int\limits_0^\Pi {\frac{{x\sin x}}{{1 + {{\cos }^2}x}}} = \Pi /2\int\limits_0^\Pi {\frac{{\sin x}}{{1 + {{\cos }^2}x}}}$

Put cosx = t
-sinx dx=dt
When $x = \Pi$ , t=-1
x=0,t=1

$\Rightarrow \int\limits_0^\Pi {\frac{{x\sin x}}{{1 + {{\cos }^2}x}}dx} = - \Pi /2\int\limits_1^{ - 1} {\frac{{dt}}{{1 + {t^2}}}}$

$\Rightarrow \int\limits_0^\Pi {\frac{{x\sin x}}{{1 + {{\cos }^2}x}}dx} = \Pi /2\int\limits_{ - 1}^1 {\frac{{dt}}{{1 + {t^2}}}}$

$\Rightarrow \int\limits_0^\Pi {\frac{{x\sin x}}{{1 + {{\cos }^2}x}}dx} = \Pi /2\left[ {{{\tan }^{ - 1}}t} \right]_{ - 1}^1$

$\Rightarrow \int\limits_0^\Pi {\frac{{x\sin x}}{{1 + {{\cos }^2}x}}dx} = \Pi /2\left[ {\frac{\Pi }{4} + \frac{\Pi }{4}} \right]$

$\Rightarrow \int\limits_0^\Pi {\frac{{x\sin x}}{{1 + {{\cos }^2}x}}dx}$ = ${\Pi ^2}/4$

Solved IGNOU Assignments and Question papers

Wednesday, June 5, 2019

IGNOU MTE-01 2019 SOLVED ASSIGNMENT 3(b)

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