Question

If $\int f\frac{{\sin }^3\left(\theta /2\right)}{\cos \left(\theta /2\right)\sqrt{{\cos }^3\theta +{\cos }^2\theta +\cos \theta }}d\theta$ then I equals

• A. ${\cot }^{-1}(\tan \theta +\sec \theta )+Const$
• B. ${\cot }^{-1}(\cos \theta +\sec \theta +1)+Const$
• C. ${\tan }^{-1}(\tan \frac{\theta }{2}+\sec \frac{\theta }{2}+1)+Const$
• D. ${\tan }^{-1}(\cos \theta +\sec \theta +1)+Const$

Answer 1

The correct option is D ${\tan }^{-1}(\cos \theta +\sec \theta +1)+Const$

Let $I=\int \frac{{\sin }^3\left(\theta /2\right)}{\cos \left(\theta /2\right)\sqrt{{\cos }^3\theta +{\cos }^2\theta +\cos \theta }}d\theta$
$=\int \frac{\sin \left(\theta /2\right)\cos \left(\theta /2\right){\sin }^2\left(\theta /2\right)}{\left({\cos }^2\left(\theta /2\right)\right)\sqrt{{\cos }^3\theta +{\cos }^2\theta +\cos \theta }}d\theta$
$=\frac{1}{2}\int \frac{\sin \theta (1-\cos \theta )}{(1+\cos \theta )\sqrt{{\cos }^3\theta +{\cos }^2\theta +\cos \theta }}d\theta$
Substitute $\cos \theta =x$
$I=-\frac{1}{2}\int \frac{(1-x)}{(1+x)\sqrt{x^3+x^2+x}}dx$
$=\frac{1}{2}\int \frac{x^2-1}{{(x+1)}^{2}x\sqrt{x+\frac{1}{x}+1}}dx$
$=\frac{1}{2}\int \frac{x^2-1}{x^2(x+\frac{1}{x}+2)\sqrt{x+\frac{1}{x}+1}}dx$
Substitute $x+\frac{1}{x}+1=t^2\Rightarrow (1-\frac{1}{x^2})dx=2tdt$
$I=\frac{1}{2}\int \frac{2tdt}{(t^2+1)t}={\tan }^{-1}t+c$
$={\tan }^{-1}(\cos \theta +\sec \theta +1)+c$