Indefinite Integration
Integration of Trigonometric Functions
Grade 12
Question:
<p>[JEE Main 2020] \(\displaystyle\int\frac{\sin x+\cos x}{\sqrt{1-\sin 2x}}\,dx\) for \(0 < x < \pi/2\)</p>
<li>\(-\ln|\cos x-\sin x|+C\)</li>
<li>\(\ln|\sin x-\cos x|+C\)</li>
<li>\(-\sin^{-1}(\sin x-\cos x)+C\)</li>
<li>\(2\ln|\cos x+\sin x|+C\)</li>
Step-by-Step Solution
Key Concept: Note 1-sin2x=(sinx-cosx)^2. For 0<x<\pi/2 near 0, cosx>sinx, so \sqrt{1-sin2x}=|cosx-sinx|=cosx-sinx. Integrand = (sinx+cosx)/(cosx-sinx).
<p>\(1-\sin 2x = (\sin x-\cos x)^2\Rightarrow\sqrt{1-\sin 2x}=|\cos x-\sin x|\).</p>
<p>For \(0<x<\pi/4\): \(\cos x>\sin x\), so \(\sqrt{1-\sin 2x}=\cos x-\sin x\).</p>
<p>\[I = \int\frac{\sin x+\cos x}{\cos x-\sin x}\,dx\]</p>
<p>Let \(u=\cos x-\sin x\Rightarrow du=-(\sin x+\cos x)\,dx\):</p>
<p>\[I = -\int\frac{du}{u} = -\ln|u|+C = -\ln|\cos x-\sin x|+C\]</p>
<p>Answer: <strong>(A)</strong></p>
Correct Answer: A