Answer:
Option A,B
Explanation:
Here, $f(x)= 7\tan^{8}x+7\tan^{6}x-3\tan^{4}x-3\tan^{2}x $
For all x $(\frac{-\pi}{2},\frac{\pi}{2})$
$\therefore$ $f(x)= 7\tan^{6}x \sec^{2}x-3\tan^{2}x \sec^{2}x$
= $(7\tan^{6}x-3\tan^{2}x)\sec^{2}x$
Now, $\int_{0}^{\frac{\pi}{4}} x f(x) dx=\int_{0}^{\frac{\pi}{4}} x(7\tan^{6}x-3\tan^{2}x)\sec^{2}x dx$
= $\left[ x(\tan^{7}x-\tan^{3}x)\right]_0^{\pi/4}$
- $\int_{0}^{\pi/4} 1(tan^{7}x-\tan^{3}x)dx$
= 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 $\Rightarrow \sec^{2}x dx=dt$
$\therefore$ $ \int_{0}^{\pi/4}x f(x)=-\int_{0}^{1} t^{3}(t^{2}-1)dt$
$=\int_{0}^{1} (t^{b}-t^{5})dt=\left[\frac{t^{4}}{4}-\frac{t^{5}}{5}\right]_0^1=\frac{1}{4}-\frac{1}{6}=\frac{1}{12}$
Also,
$\int_{0}^{\pi/4} f(x) dx= \int_{0}^{\pi/4} (7\tan^{6}x-3\tan^{2}x)\sec^{2}x dx$
= $\int_{0}^{1} (7t^{6}-3t^{2}) dt=\left[t^{7}-t^{3}\right]_0^1=0$
-
