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

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

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

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

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

Solution steps

\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 )}

= \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

= \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) ∣

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

Add a constant to the solution

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

f^{^{'}} (x) = x^{4} + 3 x^{3} + 5 x^{2} + 2 x

\int cos (πn x) d x

\frac{d y}{d x} = 4 cos (x) (3 + 2 y)^{2}

\frac{d}{d x} (\frac{x + 1}{x - 1})

\frac{d}{d x} (\frac{60}{0.01 x ^{2} + 1})

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