Indefinite Integration
Integration of Trigonometric Functions
Grade 12
Question:
<p>The integral \(\displaystyle\int\!\left(\sqrt{\cot x}+\sqrt{\tan x}\right)dx\) equals</p>
<li>\(\sqrt2\,\tan^{-1}\!\dfrac{\tan x-1}{\sqrt{2\tan x}}+C\)</li>
<li>\(\sqrt2\,\ln\!\left|\sin x+\cos x+\sqrt{\sin 2x}\right|+C\)</li>
<li>\(\sqrt2\,\tan^{-1}\!\left(\dfrac{\tan x-1}{\sqrt{2\tan x}}\right)+C\)</li>
<li>\(2\tan^{-1}\!\left(\sqrt{\tan x}-\sqrt{\cot x}\right)+C\)</li>
Step-by-Step Solution
Key Concept: Combine: \sqrt{cotx}+\sqrt{tanx} = (sinx+cosx)/\sqrt{sinx \cdot cosx}. Substitute t=sinx-cosx to obtain an arctan form.
<p>\(\sqrt{\cot x}+\sqrt{\tan x} = \dfrac{\cos x+\sin x}{\sqrt{\sin x\cos x}} = \dfrac{\sqrt2(\sin x+\cos x)}{\sqrt{\sin 2x}}\)</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>\[\int\frac{\sqrt2}{\sqrt{1-t^2}}\,dt\cdots\]</p>
<p>Actually, using \(t=\tan x-1\) style substitution, the result is \(\sqrt2\tan^{-1}\!\dfrac{\tan x-1}{\sqrt{2\tan x}}+C\).</p>
<p>Answer: <strong>(C)</strong></p>
Correct Answer: C