JEE Main 2025 Jan 29 Shift 1, Mathematics Q16: Methods of Integration

Rewrite the denominator using $\sin 2\theta =1-(\sin \theta -\cos \theta {)}^2.$ Then $9+16\sin 2\theta =25-16(\sin \theta -\cos \theta {)}^2.$

Substitute $t=\sin \theta -\cos \theta$, so $dt=(\cos \theta +\sin \theta )d\theta .$ Limits: $\theta =0\Rightarrow t=-1$; $\theta =\pi /4\Rightarrow t=0.$

$I=80{\int }_{-1}^0\frac{dt}{25-16t^2}=\frac{80}{16}{\int }_{-1}^0\frac{dt}{(5/4{)}^2-t^2}.$

Using $\int \frac{dt}{a^2-t^2}=\frac{1}{2a}\ln \left|\frac{a+t}{a-t}\right|$ with $a=5/4$:

$I=5\cdot \frac{1}{2(5/4)}{\left[\ln \left|\frac{5/4+t}{5/4-t}\right|\right]}_{-1}^0=2\left[\ln 1-\ln \frac{1/4}{9/4}\right]=-2\ln \frac{1}{9}=2\ln 9=4\ln 3.$