Question

$\int \frac{\cos \alpha +\cos x+1}{\cos \alpha +\cos x}dx=af\left(x\right)+bg\left(x\right)+c,\alpha ϵ(0,\pi )$

• A. $a=-\cos \alpha ,b=\sin \alpha ,f\left(x\right)=x+1,g\left(x\right)=\ln \left|\frac{\tan \frac{x}{2}-\cot \frac{\alpha }{2}}{\tan \frac{x}{2}+\cot \frac{\alpha }{2}}\right|$
• B. $a=\sin \alpha ,b={\cos }^2\alpha ,f\left(x\right)=\sin x,g\left(x\right)={\tan }^{-1}(\tan \frac{x}{2}+\tan \frac{\alpha }{2})$
• C. $a={\sin }^2\alpha ,b=-\cos \alpha ,f\left(x\right)=\cos x,g\left(x\right)={\tan }^{-1}\left(\frac{\tan \frac{x}{2}+\cot \frac{\alpha }{2}}{\tan \frac{x}{2}-\cot \frac{\alpha }{2}}\right)$
• D. $a=1,b=cosec\alpha ,f\left(x\right)=x,g\left(x\right)=\ln \left|\frac{\tan \frac{x}{2}+\cot \frac{\alpha }{2}}{\tan \frac{x}{2}-\cot \frac{\alpha }{2}}\right|$

Answer 1

The correct option is D $a=1,b=cosec\alpha ,f\left(x\right)=x,g\left(x\right)=\ln \left|\frac{\tan \frac{x}{2}+\cot \frac{\alpha }{2}}{\tan \frac{x}{2}-\cot \frac{\alpha }{2}}\right|$

Let $I=\int \frac{\cos \alpha +\cos x+1}{\cos \alpha +\cos x}dx$
Substitute $u=\tan \frac{x}{2}\Rightarrow du=\frac{1}{2}{\sec }^2\frac{x}{2}dx$
$I=\int \frac{2\left(\cos \alpha +\frac{1-u^2}{u^2+1}+1\right)}{(u^2+1)\left(\cos \alpha +\frac{1-u^2}{u^2+1}\right)}du$
$=2\int \frac{u^2\cos \alpha +\cos \alpha +2}{(u^2+1)(u^2\cos \alpha +\cos \alpha -u^2+1)}du$
$=2\int \frac{1}{u^2(\cos \alpha -1)+\cos \alpha +1}du+2\int \frac{1}{u^2+1}du$
$=2\int \frac{1}{(\cos \alpha +1)(\frac{u^2(\cos \alpha -1)}{\cos \alpha +1}+1)}+2\int \frac{1}{u^2+1}du$
$=\frac{1}{\cos \alpha +1}\left(\frac{{\tan }^{-1}\left(u\sqrt{-{\tan }^2\frac{a}{2}}\right)}{\sqrt{-{\tan }^2\frac{a}{2}}}\right)+2{\tan }^{-1}\left(u\right)$
$=co\sec \alpha \left(\log \left|\frac{\tan \frac{x}{2}+\cot \frac{a}{2}}{\tan \frac{x}{2}-\cot \frac{a}{2}}\right|\right)+x$