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)

$J+K=\int \frac{1+\sin x+\cos x}{1+\sin x+\cos x}dx$
$J+K=x+C$(1) $\Rightarrow$(C)
again $J-K=\int \frac{\left({\sin }^{2}x-{\cos }^{2}x\right)+\sin x-\cos x}{1+\sin x+\cos x}dx=\int \frac{(\sin x-\cos x)(\sin x+\cos x+1)}{1+\sin x+\cos x}dx$
$J-K=-\cos x-\sin x+C$....(2)
hence $J=K-(\sin x+\cos x)+C\Rightarrow$(B)
Also (1) $+(2)$
$2J=x-(\cos x+\sin x)+C$
$J=\frac{1}{2}[x-\sin x-\cos x]+C$
and $(1)-(2)$
$2K=x+(\sin x+\cos x)+C$
$K=\frac{1}{2}(x+\sin x+\cos x)+C$
from (1), $J=x-K+C\Rightarrow$ (C)