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

Answer

A, B

Explanation

Here , f (x) = 7 `=7 tan^(8)x+7tan^(6)x-3tan^(4)x-3tan^(2)x` for all `x in ((-pi)/(2),(pi)/(2))`
`:. f(x) = 7 tan^(6) x sec^(2)x -3tan^(2) sec^(2)x`
`=(7tan^(6)x-3tan^(2)x)sec^(2)x`
Now , `int_(0)^(pi//4)x f (x) dx = int_(0)^(pi//4)underset(I)(x)(7tan^(6)underset(II)(x-3)tan^(2)x)sec^(2)xdx`
`=[x(tan^(7)x-tan^(3)x)]_(0)^(pi//4)`
`-int_(0)^(pi//4)1 (tan^(7)x-tan^(3)x)dx`
`=0-int_(0)^(pi//4)tan^(3)x(tan^(4)x-1)dx`
`=-int_(0)^(pi//4)tan^(3)x(tan^(2)x-1)sec^(2)xdx`
Put tan `x=t rArrsec^(2)x dx = dt`
`:.int_(0)^(pi//4)x f (x) dx =- int_(0)^(1)t^(3)(t^(2)-1)dt`
`=int_(0)^(1)(t^(3)-t^(5))dt = [(t^(4))/(4)-(t^(5))/(5)]_(0)^(1)=(1)/(4)-(1)/(6)=(1)/(12)`
Also , `int_(0)^(pi//4)f(x)dx=int_(0)^(pi//4)(7 tan^(6)x-3tan^(2)x)sec^(2)x dx`
`=int_(0)^(1)(4t^(6)-3t^(2))dt = [t^(7)-t^(3)]_(0)^(1)=0`