JEE Main 2025 · Mathematics · Indefinite and Definite Integrals · hard

The value of ∫0^π/2(sin ^2x)/(sin x+cos x) dx is

Single correct option

The value of

\displaystyle\int_0^{\pi/2}\dfrac{\sin^2x}{\sin x+\cos x}\,dx

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Answer: (D) $\dfrac{1}{\sqrt2}\ln\left(1+\sqrt2\right)$

Let

I=\int_0^{\pi/2}\dfrac{\sin^2x}{\sin x+\cos x}dx

. Replacing

x

\dfrac\pi2-x

gives

I=\int_0^{\pi/2}\dfrac{\cos^2x}{\sin x+\cos x}dx

. Adding,

2I=\int_0^{\pi/2}\dfrac{dx}{\sin x+\cos x}

. Write

\sin x+\cos x=\sqrt2\sin\left(x+\dfrac\pi4\right)

, so

2I=\dfrac1{\sqrt2}\int_{\pi/4}^{3\pi/4}\csc u\,du=\dfrac1{\sqrt2}\left[\ln\tan\dfrac u2\right]_{\pi/4}^{3\pi/4}=\dfrac1{\sqrt2}\ln\dfrac{\tan(3\pi/8)}{\tan(\pi/8)}

. With

\tan\dfrac{3\pi}8=\sqrt2+1

and

\tan\dfrac\pi8=\sqrt2-1

the ratio is

(\sqrt2+1)^2

, so

2I=\dfrac{2}{\sqrt2}\ln(1+\sqrt2)

and

I=\dfrac1{\sqrt2}\ln(1+\sqrt2)\approx0.623

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