JEE Main 2025 — Mathematics Calculus

$\displaystyle \begin{aligned} f(x) & =7 \tan ^8 x+7 \tan ^6 x-3 \tan ^4 x-3 \tan ^2 x \\ & =7 \tan ^6 x\left(1+\tan ^2 x\right)-3 \tan ^2 x\left(1+\tan ^2 x\right) \\ & =\left(7 \tan ^6 x-3 \tan ^2 x\right)\left(1+\tan ^2 x\right) \\ & =\left(7 \tan ^6 x-3 \tan ^2 x\right) \sec ^2 x \\ l_1= & \int_0^{\frac{\pi}{4}} f(x) d x=\int_0^{\frac{\pi}{4}}\left(7 \tan ^6 x-3 \tan ^2 x\right) \sec ^2 x d x \\ & =\left.\left(\frac{7 \tan ^7 x}{7}-\frac{3 \tan ^3 x}{3}\right)\right|_0 ^{\frac{\pi}{4}}=1-1=0\end{aligned}$f(x)l1​=​=7tan8x+7tan6x−3tan4x−3tan2x=7tan6x(1+tan2x)−3tan2x(1+tan2x)=(7tan6x−3tan2x)(1+tan2x)=(7tan6x−3tan2x)sec2x∫04π​​f(x)dx=∫04π​​(7tan6x−3tan2x)sec2xdx=(77tan7x​−33tan3x​)​04π​​=1−1=0​ $\begin{aligned}
I_2= & \int_0^{\frac{\pi}{4}} x f(x) d x=\int_0^{\frac{\pi}{4}} x\left(7 \tan ^6 x-3 \tan ^2 x\right) \sec ^2 x d x \
& =\left.x\left(\tan ^7 x-\tan ^3 x\right)\right|_0 ^{\frac{\pi}{4}}-\int_0^{\frac{\pi}{4}} 1 \cdot\left(\tan ^7 x-\tan ^3 x\right) d x \
& =0-\int_0^{\frac{\pi}{4}} \tan ^3 x\left(\tan ^2 x-1\right)\left(\tan ^2 x+1\right) d x \
& =\int_0^{\frac{\pi}{4}}\left(\tan ^3 x-\tan ^5 x\right) \sec ^2 x d x=\frac{\tan ^4 x}{4}-\left.\frac{\tan ^6 x}{6}\right|_0 ^{\frac{\pi}{4}} \
& =\frac{1}{12}
\end{aligned}$Hence$Hence7 I_1+12 I_2=1$