JEE Main 2026 — Definite Integration Question with Solution

Given the function:

$f(x)={\int }_0^x\tan (t-x)dt-{\int }_0^{x}f(t)\tan tdt$

First, we simplify the first integral. Let $u=x-t$, then $du=-dt$. When $t=0$, $u=x$; when $t=x$, $u=0$.

${\int }_0^x\tan (t-x)dt={\int }_x^0\tan (-u)(-du)={\int }_x^0\tan udu=-{\int }_0^x\tan udu=-[\ln (\sec u){]}_0^x=-\ln (\sec x)=\ln (\cos x)$

So, the equation becomes:

$f(x)=\ln (\cos x)-{\int }_0^{x}f(t)\tan tdt$

Differentiating both sides with respect to $x$ using the Leibniz rule:

$f^{\prime }(x)=\frac{-\sin x}{\cos x}-f(x)\tan x$

$f^{\prime }(x)+f(x)\tan x=-\tan x$

This is a linear first-order differential equation. The integrating factor (IF) is:

$\text{IF}=e^{\int \tan xdx}=e^{\ln (\sec x)}=\sec x$

Multiplying the differential equation by $\sec x$:

$f^{\prime }(x)\sec x+f(x)\sec x\tan x=-\sec x\tan x$

$\frac{d}{dx}(f(x)\sec x)=-\sec x\tan x$

Integrating both sides with respect to $x$:

$f(x)\sec x=-\sec x+C$

To find $C$, we use the initial condition. From $f(x)=\ln (\cos x)-{\int }_0^{x}f(t)\tan tdt$, substituting $x=0$ gives $f(0)=\ln (1)-0=0$.

$0\cdot \sec 0=-\sec 0+C\Rightarrow 0=-1+C\Rightarrow C=1$

Thus, $f(x)\sec x=1-\sec x$

$f(x)=\cos x-1$

Now, we find the required derivatives:

$f^{\prime }(x)=-\sin x$

$f^{\prime \prime }(x)=-\cos x$

Evaluating these at the given points:

$f\left(\frac{\pi }{6}\right)=\cos \left(\frac{\pi }{6}\right)-1=\frac{\sqrt{3}}{2}-1$

$f^{\prime }\left(-\frac{\pi }{6}\right)=-\sin \left(-\frac{\pi }{6}\right)=\sin \left(\frac{\pi }{6}\right)=\frac{1}{2}$

$f^{\prime \prime }\left(\frac{\pi }{6}\right)=-\cos \left(\frac{\pi }{6}\right)=-\frac{\sqrt{3}}{2}$

Finally, substituting these values into the required expression:

$f^{\prime \prime }\left(\frac{\pi }{6}\right)+12f^{\prime }\left(-\frac{\pi }{6}\right)+f\left(\frac{\pi }{6}\right)=-\frac{\sqrt{3}}{2}+12\left(\frac{1}{2}\right)+\left(\frac{\sqrt{3}}{2}-1\right)$

$=-\frac{\sqrt{3}}{2}+6+\frac{\sqrt{3}}{2}-1=5$

Answer: $5$