What is the integral of 5(cos(x)+sin(x))/(sin(2x)) ?

The integral of 5(cos(x)+sin(x))/(sin(2x)) is 5(1/2 ln|tan(x/2)|+1/2 ln|tan(x)+sec(x)|)+C

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integral of 5(cos(x)+sin(x))/(sin(2x))

\int 5 \frac{cos ( x ) + sin ( x )}{sin ( 2 x )} d x

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

Solution steps

\int 5 \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

= 5 \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 )}

= 5 \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

= 5 (\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) ∣

= 5 (\frac{1}{2} ln tan (\frac{x}{2}) + \frac{1}{2} ln ∣ tan (x) + sec (x) ∣)

Add a constant to the solution

= 5 (\frac{1}{2} ln tan (\frac{x}{2}) + \frac{1}{2} ln ∣ tan (x) + sec (x) ∣) + C

f (x) = \frac{x}{x ^{2} + 81}

\int_{0}^{1} (1 - x^{3}) d x

\frac{d}{d x} ((tan (2 x))^{3})

y^{^{'}} = 1 - y^{2}

(2 x - 7)^{4} (x^{2} + x + 1)^{5}

What is the integral of 5(cos(x)+sin(x))/(sin(2x)) ?
The integral of 5(cos(x)+sin(x))/(sin(2x)) is 5(1/2 ln|tan(x/2)|+1/2 ln|tan(x)+sec(x)|)+C