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Question

Let f(x)=7tan8x+7tan6x3tan4x3tan2x for all x(π2,π2). Then the correct expression(s) is(are)

A
π40xf(x)dx=112
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B
π40f(x)dx=0
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C
π40xf(x)dx=16
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D
π40f(x)dx=1
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Solution

The correct options are
A π40xf(x)dx=112
B π40f(x)dx=0
π40f(x)dx=π40(7tan6x(sec2x1)+7tan6x3tan2x(sec2x1)3tan2x)dx
=π40(7tan6x.sec2x3tan2x.sec2x)dx
Let tanx=t sec2xdx=dt
=10(7t63t2)dt
π40f(x)dx=[t7]10[t3]10=11=0
Also, π40x.f(x)dx=π40(x.7tan6x.sec2xπ40x.3tan2x.sec2x)dx
=[xtan7x]π40π40tan7xdx[xtan3x]π40+π40tan3xdx
=π4π40tan3x(tan4x1)dxπ4
=π40tan3x(tan2x1)sec2xdx
Let tanx=t sec2xdx=dt
=10(t5t3)dt=[1614]=112

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