Let `f(x)=7tan^8x+7tan^6x-3tan^4x-3tan^4x-3tan^2x` for all `x in (-pi/2,pi/2)` . Then the correct expression (s) is (are) `int_0^(pi/4)xf(x)dx=1/(12)` `int_0^(pi/4)f(x)dx=0` `int_0^(pi/4)xf(x)=1/6` (d) `int_0^(pi/4)f(x)dx=1/(12)`

Options

A. `int_(0)^(pi//4)xf(x)dx=1/12`
B. `int_(0)^(pi//4)f(x)dx=0`
C. `int_(0)^(pi//4)xf(x)=1/6`
D. `int_(0)^(pi//4)f(x)dx=1`

Answer

A, B

Explanation

`f(x)=(7tan^(6)x-3tan^(2)x).sec^(2)x`
`:.int_(0)^((pi)/4)f(x)dx=int_(0)^(1)(7t^(6)-3t^(2))dt=(t^(7)-t^(3))_(0)^(1)=0`
Now `int_(0)^((pi)/4)xf(x)dx=int-(0)^(1)(7t^(6)-3t^(2))tan^(-1)tdt`
`=(tan^(-1)t.(t^(7)-t^(3)))_(0)^(1)-int_(0)^(1)-int_(0)^(1)(t^(7)-t^(3))1/(1+t^(2))dt`
`=int_(0)^(1)(t^(3)(1-t^(4)))/(1+t^(2))dt=int_(0)^(1)t^(3)(1-t^(2))dt=1/4-1/6=1/1`