Question

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

• A. $\frac{1}{2}{e}^x\cos ecx-x+c$
• B. $-\frac{1}{2}{e}^x\cos ecx+c$
• C. $\frac{1}{2}{e}^x\sec x+c$
• D. $-\frac{1}{2}{e}^x\sec x+c$

Answer 1

The correct option is B $-\frac{1}{2}{e}^x\cos ecx+c$

$\int e^x\left(\frac{cosx-sinx}{1-cos2x}\right)dx$

$\int e^x\left(\frac{cosx-sinx}{2si{n}^{2}x}\right)dx$

$\frac{-1}{2}\int e^x(cosecx-cotxcosecx)dx$

in the form of $\int e^x\left[f\right(x)+f^1(x\left)\right]dx=e^{x}f\left(x\right)+c$

$=\frac{-1}{2}{e}^{x}cosecx+c$

$=\frac{-1}{2}{e}^{x}cosecx+c$