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

Options

A. `overset(pi//4)underset(0)intxf(x)dx=(1)/(12)`
B. `overset(pi//4)underset(0)intxf(x)dx=0`
C. `overset(pi//4)underset(0)intxf(x)dx=(1)/(6)`
D. `overset(pi//4)underset(0)intf(x)dx=1`

Answer

A,B

Explanation

We have,
`f(x)=7 tan^(6)x(1+tan^(2)x)-3 tan^(2)x(1+tan^(2)x)`
`=(7 tan^(6)x-3 tan^(2)x)sec^(2)x`
`:.overset(pi//4)underset(0)intf(x)dx=overset(pi//4)underset(0)int (7 tan^(6)x-3tan^(2)x)sec^(2)x dx`
`rArr overset(pi//4)underset(0)int f(x)dx=overset(1)underset(0)int (7t^(6)-3t^(2))dt`, where t=tan x
`rArr overset(pi//4)underset(0)f(x)dx=[t^(7)-t^(3)]_(0)^(1)=0`.
So, option (b) is correct
`overset(pi//4)underset(0)int underset(1)(x)f underset(II)((x))dx`
`=[x(tan^(7)x-tan^(3)x)]_(0)^(pi//4)-overset(pi//4)underset(0)int(tan^(7)x-tan^(3)x)dx`
`=overset(pi//4)underset(0)int tan^(3)x(tan^(4)x-1)dx`
`=-overset(pi//4)underset(0)int tan^(3)x(tan^(4)x-1)(tan^(2)x+1)dx`
`-overset(1)underset(0)int tan^(3)x(tan^(2)x-1)sec^(2)x dx`
`=-overset(1)underset(0)int t^(3)(t^(2)-1)dt`, where t=tanx.
`=-[(t^(6))/(6)-(t^(4))/(4)]_(0)^(1)=(1)/(12)`
Hence, options(a)and (b) are true.