HSC Science (Computer Science) 12th Board ExamMaharashtra State Board
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# Solution - Evaluate :∫π0 (xsinx)/(1+sinx)dx - HSC Science (Computer Science) 12th Board Exam - Mathematics and Statistics

#### Question

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

#### 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

Is there an error in this question or solution?

#### APPEARS IN

2014-2015 (March) (with solutions)
Question 5.2.3 | 4 marks

#### Reference Material

Solution for question: Evaluate :∫π0 (xsinx)/(1+sinx)dx concept: Properties of Definite Integrals. For the courses HSC Science (Computer Science), HSC Science (General) , HSC Science (Electronics), HSC Arts, HSC Commerce, HSC Commerce (Marketing and Salesmanship)
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