What is the integral of 9(cos(x)+sin(x))/(sin(2x)) ?

The integral of 9(cos(x)+sin(x))/(sin(2x)) is 9(1/2 ln|tan(x/2)|+1/2 ln|tan(x)+sec(x)|)+C

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integral of 9(cos(x)+sin(x))/(sin(2x))

\int 9 \frac{cos ( x ) + sin ( x )}{sin ( 2 x )} d x

9 (\frac{1}{2} ln tan (\frac{x}{2}) + \frac{1}{2} ln ∣ tan (x) + sec (x) ∣) + C

Solution steps

\int 9 \cdot \frac{cos ( x ) + sin ( x )}{sin ( 2 x )} d x

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 9 \cdot \int \frac{cos ( x ) + sin ( x )}{sin ( 2 x )} d x

Expand

\frac{cos ( x ) + sin ( x )}{sin ( 2 x )} : \frac{cos ( x )}{sin ( 2 x )} + \frac{sin ( x )}{sin ( 2 x )}

= 9 \cdot \int \frac{cos ( x )}{sin ( 2 x )} + \frac{sin ( x )}{sin ( 2 x )} d x

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= 9 (\int \frac{cos ( x )}{sin ( 2 x )} d x + \int \frac{sin ( x )}{sin ( 2 x )} d x)

\int \frac{cos ( x )}{sin ( 2 x )} d x = \frac{1}{2} ln tan (\frac{x}{2})

\int \frac{sin ( x )}{sin ( 2 x )} d x = \frac{1}{2} ln ∣ tan (x) + sec (x) ∣

= 9 (\frac{1}{2} ln tan (\frac{x}{2}) + \frac{1}{2} ln ∣ tan (x) + sec (x) ∣)

Add a constant to the solution

= 9 (\frac{1}{2} ln tan (\frac{x}{2}) + \frac{1}{2} ln ∣ tan (x) + sec (x) ∣) + C

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What is the integral of 9(cos(x)+sin(x))/(sin(2x)) ?
The integral of 9(cos(x)+sin(x))/(sin(2x)) is 9(1/2 ln|tan(x/2)|+1/2 ln|tan(x)+sec(x)|)+C