JEE Advanced 2019 · Paper 2, Mathematics question 13

Symmetrise. Replacing $\theta$ by $\frac\pi2-\theta$ swaps sin and cos, so
$$2I=3\int_0^{\pi/2}\frac{d\theta}{(\sqrt{\cos\theta}+\sqrt{\sin\theta})^4}.$$

Substitute. Divide top and bottom by $\cos^2\theta$ and put $t=\tan\theta=u^2$:
$$\int_0^{\pi/2}\frac{\sec^2\theta\,d\theta}{(1+\sqrt{\tan\theta})^4}=\int_0^\infty\frac{2u\,du}{(1+u)^4}=2\left(\frac12-\frac13\right)=\frac13.$$

Value. $2I=3\times\frac13=1$, so $I=\frac12$.

Answer: 0.5