Indefinite Integration
Integration of Trigonometric Functions
Grade None
Question:
<p>[JEE Main 2022] \(\displaystyle\int\frac{\sqrt{\tan x}+\sqrt{\cot x}}{\sqrt{\sin x}\cdot\sqrt{\cos x}}\,dx\) equals (where \(C\) is a constant)</p>
<li>\(2\sqrt{2}\,\sin^{-1}(\sin x-\cos x)+C\)</li>
<li>\(2\sqrt{2}\,\cos^{-1}(\sin x+\cos x)+C\)</li>
<li>\(2\sqrt{2}\,\sin^{-1}(\cos x-\sin x)+C\)</li>
<li>\(2\tan^{-1}(\sin x-\cos x)+C\)</li>
Step-by-Step Solution
Key Concept: Simplify: (\sqrt{tanx}+\sqrt{cotx})/\sqrt{sinx cosx} = (sinx+cosx)/(sinx cosx)^(3/2) \cdot \sqrt{sinx cosx}... or note =\sqrt{2}/(sinx cosx)^(1/2) \cdot (sinx+cosx)/\sqrt{sin2x}. Let t=sinx-cosx.
<p>$\dfrac{\sqrt{\tan x}+\sqrt{\cot x}}{\sqrt{\sin x\cos x}} = \dfrac{\sqrt{\sin x}/\sqrt{\cos x}+\sqrt{\cos x}/\sqrt{\sin x}}{\sqrt{\sin x\cos x}} = \dfrac{\sin x+\cos x}{\sin x\cos x\cdot\sqrt{\sin x\cos x}}\cdot\sqrt{\sin x\cos x}$</p>
<p>Simplify: $= \dfrac{\sin x+\cos x}{\sin x\cos x}=(\sin x+\cos x)\cdot\dfrac{2}{\sin 2x}$.</p>
<p>Let $t=\sin x-\cos x\Rightarrow dt=(\cos x+\sin x)\,dx$, $t^2=1-\sin 2x\Rightarrow\sin 2x=1-t^2$.</p>
<p>$$I = 2\int\frac{dt}{1-t^2}\cdots$$ Wait, use the form with $\sqrt{1-t^2}$:</p>
<p>Actually $I = \sqrt2\int\frac{\sqrt2(\sin x+\cos x)}{\sqrt{\sin 2x}}\,dx = \sqrt2\cdot 2\int\frac{dt}{\sqrt{2-2t^2}} = 2\sqrt2\sin^{-1}(t)+C$.</p>
<p>Back-sub: $I=2\sqrt2\sin^{-1}(\sin x-\cos x)+C$. Answer: <strong>(A)</strong></p>
Correct Answer: A