Question

If $\int \frac{sinx}{1+sinx}dx=\frac{A}{798}secx-tanx+x+C$ then A is equal to

Answer 1

As $\frac{\sin x}{1+\sin x}=\frac{\sin x}{1+\sin x}\times \frac{1-\sin x}{1-\sin x}=\frac{\sin x(1-\sin x)}{{\cos }^{2}x}=\frac{\sin x-{\sin }^{2}x-{\cos }^{2}x+{\cos }^{2}x}{{\cos }^{2}x}=\frac{\sin x-1+{\cos }^{2}x}{{\cos }^{2}x}=\frac{1}{\cos x}\frac{\sin x}{\cos x}-\frac{1}{{\cos }^{2}x}+1=\sec x\tan x-{\sec }^{2}x+1$
Then
$\int \frac{\sin x}{1+\sin x}dx=\int (\sec x\tan x-{\sec }^{2}x+1)dx=\sec x-\tan x+x+c$
Therefore, $A=798$