What is the integral of (sin(x)-cos(x))/(1+sin(2x)) ?

The integral of (sin(x)-cos(x))/(1+sin(2x)) is -(sin(x)-cos(x))/(tan(x)+1)+2cos^2(x/2)tan(x/2)+C

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integral of (sin(x)-cos(x))/(1+sin(2x))

\int \frac{sin ( x ) - cos ( x )}{1 + sin ( 2 x )} d x

- \frac{sin ( x ) - cos ( x )}{tan ( x ) + 1} + 2 cos^{2} (\frac{x}{2}) tan (\frac{x}{2}) + C

Solution steps

\int \frac{sin ( x ) - cos ( x )}{1 + sin ( 2 x )} d x

Apply Integration By Parts

: - \frac{sin ( x ) - cos ( x )}{tan ( x ) + 1} - \int - \frac{cos ( x ) + sin ( x )}{tan ( x ) + 1} d x

= - \frac{sin ( x ) - cos ( x )}{tan ( x ) + 1} - \int - \frac{cos ( x ) + sin ( x )}{tan ( x ) + 1} d x

\int - \frac{cos ( x ) + sin ( x )}{tan ( x ) + 1} d x = - \frac{2 tan ( \frac{x}{2} )}{sec ^{2} ( \frac{x}{2} )}

= - \frac{sin ( x ) - cos ( x )}{tan ( x ) + 1} - (- \frac{2 tan ( \frac{x}{2} )}{sec ^{2} ( \frac{x}{2} )})

Simplify

- \frac{sin ( x ) - cos ( x )}{tan ( x ) + 1} - (- \frac{2 tan ( \frac{x}{2} )}{sec ^{2} ( \frac{x}{2} )}) : - \frac{sin ( x ) - cos ( x )}{tan ( x ) + 1} + 2 cos^{2} (\frac{x}{2}) tan (\frac{x}{2})

= - \frac{sin ( x ) - cos ( x )}{tan ( x ) + 1} + 2 cos^{2} (\frac{x}{2}) tan (\frac{x}{2})

Add a constant to the solution

= - \frac{sin ( x ) - cos ( x )}{tan ( x ) + 1} + 2 cos^{2} (\frac{x}{2}) tan (\frac{x}{2}) + C

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What is the integral of (sin(x)-cos(x))/(1+sin(2x)) ?
The integral of (sin(x)-cos(x))/(1+sin(2x)) is -(sin(x)-cos(x))/(tan(x)+1)+2cos^2(x/2)tan(x/2)+C