JEE Main 2025 Jan 22 Shift 1, Mathematics Q15: Integration by Parts

Factor: $f(x)=7{\tan }^{6}x({\tan }^{2}x+1)-3{\tan }^{2}x({\tan }^{2}x+1)=(7{\tan }^{6}x-3{\tan }^{2}x){\sec }^{2}x$.

For $I_1$, substitute $t=\tan x$, $dt={\sec }^{2}xdx$:

$I_1={\int }_0^1(7t^6-3t^2)dt={\left[t^7-t^3\right]}_0^1=0$.

For $I_2$, integrate by parts with $u=x$, $dv=(7{\tan }^{6}x-3{\tan }^{2}x){\sec }^{2}xdx$, so $v={\tan }^{7}x-{\tan }^{3}x$:

$I_2={\left[x({\tan }^{7}x-{\tan }^{3}x)\right]}_0^{\pi /4}-{\int }_0^{\pi /4}({\tan }^{7}x-{\tan }^{3}x)dx=0-{\int }_0^{\pi /4}{\tan }^{3}x({\tan }^{4}x-1)dx$.

$=-{\int }_0^{\pi /4}{\tan }^{3}x({\tan }^{2}x-1)({\tan }^{2}x+1)dx$. Substituting $t=\tan x$ gives

$I_2=-{\int }_0^1{t}^3(t^2-1)dt=-{\left[\frac{t^6}{6}-\frac{t^4}{4}\right]}_0^1=-\left(\frac{1}{6}-\frac{1}{4}\right)=\frac{1}{12}$.

$7I_1+12I_2=0+12\cdot \frac{1}{12}=1$.