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Question

# Let f(x)=7tan8x+7tan6x−3tan4x−3tan2x 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 π4∫0xf(x)dx=112 B π4∫0f(x)dx=0π4∫0f(x)dx=π4∫0(7tan6x(sec2x−1)+7tan6x−3tan2x(sec2x−1)−3tan2x)dx =π4∫0(7tan6x.sec2x−3tan2x.sec2x)dx Let tanx=t ⇒sec2xdx=dt =1∫0(7t6−3t2)dt ⇒π4∫0f(x)dx=[t7]10−[t3]10=1−1=0 Also, π4∫0x.f(x)dx=π4∫0(x.7tan6x.sec2x−π4∫0x.3tan2x.sec2x)dx =[xtan7x]π40−π4∫0tan7xdx−[xtan3x]π40+π4∫0tan3xdx =π4−π4∫0tan3x(tan4x−1)dx−π4 =−π4∫0tan3x(tan2x−1)sec2xdx Let tanx=t ⇒sec2xdx=dt =−1∫0(t5−t3)dt=−[16−14]=112

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