Mathematics - Definite Integration Question with Solution | TestHub

$f\left(x\right)=7{\tan }^{8}x+7{\tan }^{6}x-3{\tan }^{4}x-3{\tan }^{2}x$
$f\left(x\right)={\sec }^{2}x\left(7{\tan }^{6}x-3{\tan }^{2}x\right)$
Now,
$\underset{0}{\overset{\frac{\pi }{4}}{\int }}{\sec }^{2}x\left(7{\tan }^{6}x-3{\tan }^{2}x\right) dx$
$={\int }_0^1\left(7t^6-3t^2\right)dt$
$=\frac{7t^7}{7}-\frac{3t^3}{3}$
$=1^7-1^3=0$
Also
${\int }_0^{\frac{\pi }{4}}x{\sec }^{2}x \left(7{\tan }^{6}x-3{\tan }^{2}x\right) dx$
$\tan x=t$
${\int }_0^1{\tan }^{-1}t\left(7t^6-3t^2\right) dt$
${\tan }^{-1}t\left(t^7-t^3\right){|}_0^1-{\int }_0^1\frac{1}{1+t^2} \left(t^7-t^3\right)dt$
$=-{\int }_0^1\frac{1}{1+t^2} t^3\left(t^4-1\right)dt$
$=+\underset{0}{\overset{1}{\int }}\frac{t^3 \left(1+t^2\right) \left(1-t^2\right)dt}{1+t^2}=\underset{0}{\overset{1}{\int }}\left(t^3-t^5\right)dt$
$=\frac{1}{4}-\frac{1}{6}=\frac{1}{12}$