Maharashtra State BoardHSC Arts 12th Board Exam
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Evaluate :∫π0 (xsinx)/(1+sinx)dx - Mathematics and Statistics

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Evaluate :`int_0^pi(xsinx)/(1+sinx)dx`

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Solution

Let `I=int_0^pi(xsinx)/(1+sinx)dx`

`=int_0^pi((pi-x)sin(pi-x))/(1+sin(pi-x))dx [because int_0^a f(x)dx=int_0^af(a-x)dx]`

 

`=int_0^pi((pi-x)sinx)/(1+sinx)dx`

 

`=int_0^pi(pisinx)/(1+sinx)dx-I`

 

`I=int_0^pi(pisinx)/(1+sinx)dx-I`

 

`2I=int_0^pi(pisinx.(1-sinx))/((1+sinx)(1-sinx))dx`

 

`2I=int_0^pi(pisinx.(1-sinx))/(1-sin^2x)dx`

 

`(2I)/pi=int_0^pi(sinx.(1-sinx))/cos^2xdx`

 

`(2I)/pi=int_0^pi(sinx.-sin^2x)/cos^2xdx`

`(2I)/pi=int_0^pi(sinx)/cos^2xdx-int_0^pi(sin^2x)/cos^2xdx`

 

`(2I)/pi=int_0^pisecx.tanxdx-int_0^pitan^2xdx`

 

`(2I)/pi=[secx]_0^pi-int_0^pi(sec^2x-1)dx`

 

`(2I)/pi=[secpi-sec0]-int_0^pisec^2x.dx+int_0^pi1dx`

`(2I)/pi=[-1-1]-[tanx]_0^pi_[x]_0^pi`

`(2I)/pi=[-2]-[tanpi-tan0]+pi`

`(2I)/pi=[-2]-0+pi`

`thereforeI=((pi-2)pi)/2`

Concept: Properties of Definite Integrals
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