Definite Integration
Grade 12
Question:
<p>Evaluate \(\displaystyle\int_0^{\pi/2}\frac{\sin^2 x}{\sin x+\cos x}\,dx\) [JEE Main 2022]</p>
(1/\sqrt{2}) \cdot ln(\sqrt{2}+1)-1/\sqrt{2}
\pi/(4\sqrt{2})
(\sqrt{2}-1)/2
(1/\sqrt{2})ln(1+\sqrt{2})
Step-by-Step Solution
Key Concept: Write sin^2x = (1/2)(1-cos2x). King's rule to add cos^2x/(sinx+cosx). Then manipulate.
<div class='solution'>
<p>Let \(I=\int_0^{\pi/2}\frac{\sin^2 x}{\sin x+\cos x}dx\), \(J=\int_0^{\pi/2}\frac{\cos^2 x}{\sin x+\cos x}dx\).</p>
<p>By King, \(I=J\). And \(I+J=\int_0^{\pi/2}\frac{\sin^2x+\cos^2x}{\sin x+\cos x}dx=\int_0^{\pi/2}\frac{dx}{\sin x+\cos x}\).</p>
<p>\(\sin x+\cos x=\sqrt{2}\sin(x+\pi/4)\). \(\int_0^{\pi/2}\frac{dx}{\sqrt{2}\sin(x+\pi/4)}=\frac{1}{\sqrt{2}}\int_{\pi/4}^{3\pi/4}\csc u\,du=\frac{1}{\sqrt{2}}[\ln|\csc u-\cot u|]_{\pi/4}^{3\pi/4}=\frac{1}{\sqrt{2}}\ln(\sqrt{2}+1)\cdot 2\).</p>
<p>So \(2I=\frac{\sqrt{2}\ln(\sqrt{2}+1)}{\sqrt{2}}\Rightarrow I=\frac{\ln(1+\sqrt{2})}{\sqrt{2}}\).</p>
Correct Answer: A