JEE Advanced 2015 · Paper 2, Mathematics question 55

Factor.
$$f(x)=(7\tan^6x-3\tan^2x)(1+\tan^2x)=(7\tan^6x-3\tan^2x)\sec^2x.$$

(B).
$$\int_0^{\pi/4}f\,dx=\Big[\tan^7x-\tan^3x\Big]_0^{\pi/4}=0.$$
True, so (D) is false.

(A). Integrate by parts:
$$\int_0^{\pi/4}xf\,dx=\Big[x(\tan^7x-\tan^3x)\Big]_0^{\pi/4}-\int_0^{\pi/4}(\tan^7x-\tan^3x)\,dx.$$
The bracket is 0. Since $\tan^7-\tan^3=\tan^3x(\tan^2x-1)\sec^2x$, substitute $t=\tan x$:
$$-\int_0^1t^3(t^2-1)\,dt=-\left(\frac16-\frac14\right)=\frac1{12}.$$
True, so (C) is false.

(A), (B)