JEE Main 2025 — 22 January, Morning Shift — Question 15

$f(x)=\left(7 \tan ^{6} x-3 \tan ^{2} x\right)\left(\sec ^{2} x\right)$

$I_{1}=\int_{0}^{\pi / 4}\left(7 \tan ^{6} x-3 \tan ^{2} x\right)\left(\sec ^{2} x\right) d x$

Put $\tan \mathrm{x}=\mathrm{t}$

$I_{1}=\int_{0}^{1}\left(7 t^{6}-3 t^{2}\right) d t=\left[t^{7}-t^{3}\right]_{0}^{1}=0$

$I_{2}=\int_{0}^{\pi / 4} x \underbrace{\left(7 \tan ^{6} x-3 \tan ^{2} x\right)\left(\sec ^{2} x\right)}_{\mathrm{II}} d x$

$=\left[x\left(\tan ^{7} x-\tan ^{3} x\right)\right]_{0}^{\pi / 4}-\int_{0}^{\pi / 4}\left(\tan ^{7} x-\tan ^{3} x\right) d x$

$=0-\int_{0}^{\pi / 4} \tan ^{3} x\left(\tan ^{2} x-1\right)\left(1+\tan ^{2} x\right) d x$

Put $\tan \mathrm{x}=\mathrm{t}$

$=-\int_{0}^{1}\left(\mathrm{t}^{5}-\mathrm{t}^{3}\right) \mathrm{dt}=-\left[\frac{\mathrm{t}^{6}}{6}-\frac{\mathrm{t}^{4}}{4}\right]=\frac{1}{12}$

$7 \mathrm{I}_{1}+12 \mathrm{I}_{2}=1$