Mathematics - Definite Integration Question with Solution | TestHub

$\begin{array}{rl}f(x)& =7{\tan }^{8}x+7{\tan }^{6}x-3{\tan }^{4}x-3{\tan }^{2}x\\ & =7{\tan }^{6}x\left(1+{\tan }^{2}x\right)-3{\tan }^{2}x\left(1+{\tan }^{2}x\right)\\ & =\left(7{\tan }^{6}x-3{\tan }^{2}x\right)\left(1+{\tan }^{2}x\right)\\ & =\left(7{\tan }^{6}x-3{\tan }^{2}x\right){\sec }^{2}x\\ l_1=& {\int }_0^{\frac{\pi }{4}}f(x)dx={\int }_0^{\frac{\pi }{4}}\left(7{\tan }^{6}x-3{\tan }^{2}x\right){\sec }^{2}xdx\\ & ={\left(\frac{7{\tan }^{7}x}{7}-\frac{3{\tan }^{3}x}{3}\right)|}_0^{\frac{\pi }{4}}=1-1=0\end{array}$ $\begin{array}{rl}{I}_2=& {\int }_0^{\frac{\pi }{4}}xf(x)dx={\int }_0^{\frac{\pi }{4}}x\left(7{\tan }^{6}x-3{\tan }^{2}x\right){\sec }^{2}xdx\\ & ={x\left({\tan }^{7}x-{\tan }^{3}x\right)|}_0^{\frac{\pi }{4}}-{\int }_0^{\frac{\pi }{4}}1\cdot \left({\tan }^{7}x-{\tan }^{3}x\right)dx\\ & =0-{\int }_0^{\frac{\pi }{4}}{\tan }^{3}x\left({\tan }^{2}x-1\right)\left({\tan }^{2}x+1\right)dx\\ & ={\int }_0^{\frac{\pi }{4}}\left({\tan }^{3}x-{\tan }^{5}x\right){\sec }^{2}xdx=\frac{{\tan }^{4}x}{4}-{\frac{{\tan }^{6}x}{6}|}_0^{\frac{\pi }{4}}\\ & =\frac{1}{12}\end{array}$ Hence $7I_1+12I_2=1$