JEE Main 2025 — 29 January, Morning Shift — Question 60

$I=80 \int_{0}^{\frac{\pi}{4}}\left(\frac{\sin \theta+\cos \theta}{9+16(2 \sin \theta \cdot \cos \theta)}\right) d \theta$

$=80 \int_{0}^{\frac{\pi}{4}} \frac{\sin \theta+\cos \theta}{9-16(1-2 \sin \theta \cdot \cos \theta-1)} d \theta$

$=80 \int_{0}^{\frac{\pi}{4}} \frac{\sin \theta+\cos \theta}{9+16-16(\sin \theta-\cos \theta)^{2}} d \theta$

Let $\sin \theta-\cos \theta=\mathrm{t}$

$(\cos \theta+\sin \theta) d \theta=d t$

$=80 \int_{-1}^{0} \frac{\mathrm{dt}}{25-16 \mathrm{t}^{2}}$

$=\frac{80}{16} \int_{-1}^{0} \frac{\mathrm{dt}}{\left(\frac{5}{4}\right)^{2}-\mathrm{t}^{2}}$

$\left.=\frac{5}{2\left(\frac{5}{4}\right)} \ln \left(\frac{\frac{5}{4}+t}{\frac{5}{4}-t}\right)\right]_{-1}^{0}$

$=2 \ln (1)+4 \ln 3$ $=4 \ln 3$