HSC Arts 12th Board ExamMaharashtra State Board
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# Evaluate ∫0(3/2) |x cosπx| dx - HSC Arts 12th Board Exam - Mathematics and Statistics

ConceptEvaluation of Definite Integrals by Substitution

#### Question

Evaluate ∫_0^(3/2)|x cosπx|dx

#### Solution

int_0^(3/2)|xcospix|dx

0<x<1/2

0<pix<pi/2rArrcospix>0rArr(xcospix)>0

|xcospix|=xcospix

1/2<x<3/2

pi/2<pix<(3pi)/2rArrcospix<0rArr(xcospix)<0

|xcospix|=-xcospix

I=int_0^(3/2)|xcospix|dx=int_0^(3/2)xcospix+int_(1/2)^(3/2)-(xcospix)

I=int_0^(1/2)xcospix-int_(1/2)^(3/2)xcospix

intx(cospix)=x(sinpix)/pi-int(sinpix)/pi

=x/pi(sinpix)+(cospix)/pi^2

I=[(x/pisinpix)+(cospix)/pi^2]_0^(1/2)-[(x/pisinpix)+(cospix)/pi^2]_(1/2)^(3/2)

=[1/pi((1/2)-0)+1/pi^2(0-1)]-[1/pi(3/2(-1)-1/2(1))+1/pi^2(0-0)]

=(1/(2pi)-1/pi^2)-((-2)/pi)

=(5/(2pi)-1/pi^2)

=((5pi-2)/(2pi^2))

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Solution Evaluate ∫0(3/2) |x cosπx| dx Concept: Evaluation of Definite Integrals by Substitution.
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