JEE Main · 2019 · 09 Apr 2019 · Shift 1 · Definite Integrals

Call the integral $I$. Replacing $x$ by $\frac{\pi }{2}-x$ swaps sine and cosine in the numerator while leaving the denominator $\sin x+\cos x$ unchanged, so $I={\int }_0^{\pi /2}\frac{{\cos }^{3}x}{\sin x+\cos x}dx$ as well. Adding the two forms, the numerator becomes ${\sin }^{3}x+{\cos }^{3}x$, which factors over $\sin x+\cos x$ to ${\sin }^{2}x-\sin x\cos x+{\cos }^{2}x=1-\frac{1}{2}\sin 2x$. So $2I={\int }_0^{\pi /2}\left(1-\frac{1}{2}\sin 2x\right)dx={\left[x+\frac{1}{4}\cos 2x\right]}_0^{\pi /2}=\frac{\pi }{2}-\frac{1}{2}$, giving $I=\frac{\pi -1}{4}$.