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Question

The value of π2π2(x3+xcosx+tan5x+1)dx is .

A
2π
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B
6π
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C
3π
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D
4π
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Solution

The correct option is A 2π
I=π2π2(x3+xcosx+tan8x+1)dx
=⎢ ⎢π2π2x3+π2π2xcosx+π2π2tan5x+π2π21.⎥ ⎥
Let I1=π2π2x3=[x44]π/2π/2=0
I2=π2π2xcosx
By Integration by parts method
=xcosdx1.cosx dx.dx
=xsinxsinx dx
=[xsinx+cosx]π/2π/2=π2(1)+0π2(1)
I3=π/2π/2tan5x dx=π/2π/2tan3x.tan2xdx
=π/2π/2tan3x(sec2x1)dx=π/2π/2tan3xsec2x dxπ/2π/2tan3x dx
=π/2π/2tan3xsec3x dxπ/2π/2tanx.tan2x dx
=π/2π/2tan3xsec2x dxπ/2π/2π/2π/2tanx.(sec2x1)dx
=π/2π/2tan3xsinx dxπ/2π/2tanxsec2x dx+π/2π/2tanx dx
Let tanx=t dt=sec2x dx
=t3 dtt dt+ln|secx|
=[tan4x4tan2x2+ln|secx|]π/2π/2
=0
I4=π/2π/21.dx=[x]π/2π/2=π
I=I1+I2+I3+I4
=0+π+0+π
=2π


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