Question

Suppose J $=\int \frac{si{n}^{2}x+sinx}{1+sinx+cosx}dx$ and $K=\int \frac{co{s}^{2}x+cosx}{1+sinx+cosx}dx$. If C is an arbitrary constant of integration then which of the following is correct?

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

Answer 1

The correct option is B $J=K-(sinx+cosx)+C$

$J=\int \frac{{\sin }^{2}x+\sin x}{1+\sin x+\cos x}$
Put, $u=\tan \left(\frac{x}{2}\right)\Rightarrow du=\frac{1}{2}{\sec }^2\frac{x}{2}dx$
$J=\int \frac{2(\frac{{4u}^2}{{(u^2+1)}^2}+\frac{2u}{u^2+1})}{(u^2+1)(\frac{2u}{u^2+1}+\frac{1-u^2}{u^2+1}+1)}du$
$=2\int (\frac{u-1}{{(u^2+1)}^2}+\frac{1}{u^2+1})du=2{\tan }^{-1}u+2\int \frac{u-1}{{(u^2+1)}^2}du$
In the integral, put $u=\tan t\Rightarrow du=\Rightarrow {\sec }^{2}tdt$
$\therefore J=2{\tan }^{-1}u+2\int {\cos }^{2}t(\tan t-1)dt=2{\tan }^{-1}u+2\int (1-{\sin }^{2}t)(\tan t-1)=2{\tan }^{-1}u-\cos \left({\tan }^{-1}u\right)2-\sin \left({\tan }^{-1}u\right)\cos \left({\tan }^{-1}u\right)=\frac{1}{2}(x-\sin x-\cos x)$
$K=\int \frac{{\cos }^{2}x+\cos x}{1+\cos x+\sin x}dx$
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 $u=\tan t\Rightarrow du={\sec }^{2}tdt$
$\therefore K=-\int {\cos }^{2}t\left(\tan t-1\right)dt={\tan }^{-1}u+\cos {\left({\tan }^{-1}u\right)}^2+\sin \left({\tan }^{-1}u\right)\cos \left({\tan }^{-1}u\right)\therefore J=K-(\sin x+\cos x)+c$