Question

$\frac{{\sin }^{3}x}{({\cos }^{4}x+3{\cos }^{2}x+1){\tan }^{-1}(\sec x+\cos x)}dx=$

• A. ${\tan }^{-1}(\sec x+\cos x)+c$
• B. $\ln |{\tan }^{-1}(\sec x+\cos x)|+c$
• C. $\frac{1}{(\sec x+{\cos }^{2}x{)}^2}+c$
• D. $\ln |\sec x+\cos x|+c$

Answer 1

The correct option is B $\ln |{\tan }^{-1}(\sec x+\cos x)|+c$

Put ${\tan }^{-1}(\sec x+\cos x)=f\left(x\right)$
$\Rightarrow f^{\prime }\left(x\right)=\frac{d}{dx}\left({\tan }^{-1}\right(\sec x+\cos x\left)\right)$
$=\frac{1}{1+(\sec x+\cos x{)}^2}\times \frac{d}{dx}(\sec x+\cos x)$
$=\frac{1}{1+(\frac{1}{\cos x}+\cos x{)}^2}\times (\sec x\tan x-sinx)$
$=\frac{1}{1+{\cos }^{2}x+\frac{1}{{\cos }^{2}x}+2}\times (\frac{1}{\cos x}\times \frac{\sin x}{\cos x}-sinx)$

$=\frac{1}{\frac{{\cos }^{4}x+3{\cos }^x+1}{{\cos }^{2}x}}\times (\frac{\sin x}{{\cos }^{2}x}-\sin x)$

$=\frac{1}{\frac{{\cos }^{4}x+3{\cos }^x+1}{{\cos }^{2}x}}\times \left(\frac{\sin x-\sin x{\cos }^{2}x}{{\cos }^{2}x}\right)$
$=\frac{1}{{\cos }^{4}x+3{\cos }^x+1}\times \sin x(1-{\cos }^{2}x)$
$=\frac{1}{{\cos }^{4}x+3{\cos }^x+1}\times \left({\sin }^{3}x\right)$


$\Rightarrow f^{\prime }\left(x\right)=\frac{{\sin }^{3}x}{{\cos }^{4}x+3{\cos }^{2}x+1}\therefore \int \frac{f^{\prime }\left(x\right)dx}{f\left(x\right)}=\ln |f\left(x\right)|+c$

Thus the integral is $\ln |{\tan }^{-1}(\sec x+\cos x)|+c$