Let for $f(x)=7 an ^8 x+7 an ^6 x-3 an ^4 x-3 an ^2 x,$ $\mathrm{I}_1=\int_0^{\pi / 4} f(x) \mathrm{d} x$ and $\mathrm{I}_2=\int_0^{\pi / 4} x f(x) \mathrm{d} x$. Then $7 \mathrm{I}_1+12 \mathrm{I}_2$ is equal to :

Question

Let for $f(x)=7 	an ^8 x+7 	an ^6 x-3 	an ^4 x-3 	an ^2 x,$ $\mathrm{I}_1=\int_0^{\pi / 4} f(x) \mathrm{d} x$ and $\mathrm{I}_2=\int_0^{\pi / 4} x f(x) \mathrm{d} x$. Then $7 \mathrm{I}_1+12 \mathrm{I}_2$ is equal to :

Aicharya · Accepted Answer

$\begin{aligned} f(x) & =7 	an ^8 x+7 	an ^6 x-3 	an ^4 x-3 	an ^2 x \ & =7 	an ^6 x\left(1+	an ^2 xight)-3 	an ^2 x\left(1+	an ^2 xight) \ & =\left(7 	an ^6 x-3 	an ^2 xight)\left(1+	an ^2 xight) \ & =\left(7 	an ^6 x-3 	an ^2 xight) \sec ^2 x \ I_1= & \int_0^{\frac{\pi}{4}} f(x) d x=\int_0^{\frac{\pi}{4}}\left(7 	an ^6 x-3 	an ^2 xight) \sec ^2 x d x \ & =\left.\left(\frac{7 	an ^7 x}{7}-\frac{3 	an ^3 x}{3}ight)ight|_0 ^{\frac{\pi}{4}}=1-1=0\end{aligned}$ $\begin{aligned} I_2= & \int_0^{\frac{\pi}{4}} x f(x) d x=\int_0^{\frac{\pi}{4}} x\left(7 	an ^6 x-3 	an ^2 xight) \sec ^2 x d x \ & =\left.x\left(	an ^7 x-	an ^3 xight)ight|_0 ^{\frac{\pi}{4}}-\int_0^{\frac{\pi}{4}} 1 \cdot\left(	an ^7 x-	an ^3 xight) d x \ & =0-\int_0^{\frac{\pi}{4}} 	an ^3 x\left(	an ^2 x-1ight)\left(	an ^2 x+1ight) d x \ & =\int_0^{\frac{\pi}{4}}\left(	an ^3 x-	an ^5 xight) \sec ^2 x d x=\frac{	an ^4 x}{4}-\left.\frac{	an ^6 x}{6}ight|_0 ^{\frac{\pi}{4}} \ & =\frac{1}{12} \end{aligned}$ Hence $7 I_1+12 I_2=1$

Let for $f(x)=7 \tan ^8 x+7 \tan ^6 x-3 \tan ^4 x-3 \tan ^2 x,$ $\mathrm{I}_1=\int_0^{\pi / 4} f(x)…

Solution