Question

Suppose $J=\int \frac{{\sin }^{2}x+\sin x}{1+\sin x+\cos x}dx$ and $K=\int \frac{{\cos }^{2}x+\cos x}{1+\sin x+\cos x}dx$. If $C$ is an arbitrary constant of integration then which of the following is/are correct?

• A. $J=\frac{1}{2}(x-\sin x+\cos x)+C$
• B. $J=K-(\sin x+\cos x)+C$
• C. $J=x-K+C$
• D. $K=\frac{1}{2}(x-\sin x+\cos x)+C$

Answer 1

Answer: (B), (C)

$J=K-(\sin x+\cos x)+C$
$J=x-K+C$

$J=\int \frac{{\sin }^{2}x+\sin x}{1+\sin x+\cos x}$

Put $u=\tan \left(\frac{x}{2}\right)\Rightarrow du=\frac{1}{2}{\sec }^{2}xdx$

$J=\int \frac{2(\frac{{4u}^2}{{(u^2+1)}^2}+\frac{2u}{u^2+1})}{(u^2+1)(\frac{2u}{u^2+1}+\frac{1-u^2}{u^2+1}+1)}du$

$=2\int (\frac{u-1}{{(u^2+1)}^2}+\frac{1}{u^2+1})du=2{\tan }^{-1}u+2\int \frac{u-1}{{(u^2+1)}^2}du$

Put $u=\tan t\Rightarrow du=\Rightarrow {\sec }^{2}tdt$

$\therefore J=2{\tan }^{-1}u+2\int {\cos }^{2}t(\tan t-1)dt=2{\tan }^{-1}u+2\int (1-{\sin }^{2}t)(tant-1)=2{\tan }^{-1}u-\cos \left({\tan }^{-1}u\right)2-\sin \left({\tan }^{-1}u\right)\cos \left({\tan }^{-1}u\right)$

$=\frac{1}{2}(x-\sin x-\cos x)$

$K=\int \frac{{\cos }^{2}x+\cos x}{1+\cos x+\sin x}dx$

Put $u=\tan \frac{x}{2}\Rightarrow du=\frac{1}{2}{\sec }^2\frac{x}{2}dx$

$K=-\int \frac{2(u-1)}{u^2+{2u}^2+1}du$

Put $u=\tan t\Rightarrow du={\sec }^{2}tdt$

$\therefore k=-\int {\cos }^{2}t\left(\tan t-1\right)dt={\tan }^{-1}u+\cos {\left({\tan }^{-1}u\right)}^2+\sin \left({\tan }^{-1}u\right)\cos \left({\tan }^{-1}u\right)\therefore J=K-(\sin x+\cos x)+c$