Question

Suppose $J=\int \frac{\sin ^2 x+\sin x}{1+\sin x+\cos x} d x$ and $K=\int \frac{\cos ^2 x+\cos x}{1+\sin x+\cos x} d x$. 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)

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