Let for f(x)=7 ^8x+7 ^6x-3 ^4x-3 ^2x,, I_1= _ /4^0f(x)dx and I_2= _ /4^0xf(x)dx.

Question

Let for $f(x)=7{	an }^{8}x+7{	an }^{6}x-3{	an }^{4}x-3{	an }^{2}x,\, {I}_{1}={\int }_{\pi /4}^{0}f(x)dx$ and ${I}_{2}={\int }_{\pi /4}^{0}xf(x)dx$. Then $7{I}_{1}+12{I}_{2}$ is equal to :

Accepted Answer

$\begin{aligned} f(x) & =7 \tan ^8 x+7 \tan ^6 x-3 \tan ^4 x-3 \tan ^2 x \ & =7 \tan ^6 x\left(1+\tan ^2 x\right)-3 \tan ^2 x\left(1+\tan ^2 x\right) \ & =\left(7 \tan ^6 x-3 \tan ^2 x\right)\left(1+\tan ^2 x\right) \ & =\left(7 \tan ^6 x-3 \tan ^2 x\right) \sec ^2 x\end{aligned}$

$\begin{aligned} I_1= & \int_0^{\frac{\pi}{4}} f(x) d x=\int_0^{\frac{\pi}{4}}\left(7 \tan ^6 x-3 \tan ^2 x\right) \sec ^2 x d x \ & =\left.\left(\frac{7 \tan ^7 x}{7}-\frac{3 \tan ^3 x}{3}\right)\right|_0 ^{\frac{\pi}{4}}=1-1=0\end{aligned}$

$\begin{aligned} I_2= & \int_0^{\frac{\pi}{4}} x f(x) d x=\int_0^{\frac{\pi}{4}} x\left(7 \tan ^6 x-3 \tan ^2 x\right) \sec ^2 x d x \ & =\left.x\left(\tan ^7 x-\tan ^3 x\right)\right|_0 ^{\frac{\pi}{4}}-\int_0^{\frac{\pi}{4}} 1 \cdot\left(\tan ^7 x-\tan ^3 x\right) d x \ & =0-\int_0^{\frac{\pi}{4}} \tan ^3 x\left(\tan ^2 x-1\right)\left(\tan ^2 x+1\right) d x\end{aligned}$

$$ \begin{aligned} & =\int_0^{\frac{\pi}{4}}\left(\tan ^3 x-\tan ^5 x\right) \sec ^2 x d x=\frac{\tan ^4 x}{4}-\left.\frac{\tan ^6 x}{6}\right|_0 ^{\frac{\pi}{4}} \ & =\frac{1}{12} \end{aligned} $$

Hence $7 I_1+12 I_2=1$

Solutions · 1