Question

The value of $\int \frac{1}{\sec x+\tan x+\csc x+\cot x}dx$ is:

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

Answer 1

Answer: (B)

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

$Q=\int \frac{1}{\sec x+\tan x+cosecx+\cot x}dx$

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

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

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

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

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

$=\frac{1}{2}\int \frac{(\sin x+\cos x{)}^2-1^2}{(1+\sin x+\cos x)}dx$

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

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

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

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