What is the integral of (3cos(x))/(sin^2(x)+sin(x)) ?

The integral of (3cos(x))/(sin^2(x)+sin(x)) is -3(-ln|tan(x/2)|+2ln|tan(x/2)+1|)+C

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integral of (3cos(x))/(sin^2(x)+sin(x))

\int \frac{3 cos ( x )}{sin ^{2} ( x ) + sin ( x )} d x

- 3 (- ln tan (\frac{x}{2}) + 2 ln tan (\frac{x}{2}) + 1) + C

Solution steps

\int \frac{3 cos ( x )}{sin ^{2} ( x ) + sin ( x )} d x

Take the constant out:

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

= 3 \cdot \int \frac{cos ( x )}{sin ^{2} ( x ) + sin ( x )} d x

Rewrite using trig identities

= 3 \cdot \int \frac{cos ( x )}{1 - cos ^{2} ( x ) + sin ( x )} d x

Apply u-substitution

= 3 \cdot \int - \frac{u - 1}{u ( u + 1 )} d u

Take the constant out:

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

= 3 (- \int \frac{u - 1}{u ( u + 1 )} d u)

Take the partial fraction of

\frac{u - 1}{u ( u + 1 )} : - \frac{1}{u} + \frac{2}{u + 1}

= 3 (- \int - \frac{1}{u} + \frac{2}{u + 1} d u)

Apply the Sum Rule:

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

= 3 (- (- \int \frac{1}{u} d u + \int \frac{2}{u + 1} d u))

\int \frac{1}{u} d u = ln ∣ u ∣

\int \frac{2}{u + 1} d u = 2 ln ∣ u + 1 ∣

= 3 (- (- ln ∣ u ∣ + 2 ln ∣ u + 1 ∣))

Substitute back

u = tan (\frac{x}{2})

= 3 (- (- ln tan (\frac{x}{2}) + 2 ln tan (\frac{x}{2}) + 1))

Simplify

= - 3 (- ln tan (\frac{x}{2}) + 2 ln tan (\frac{x}{2}) + 1)

Add a constant to the solution

= - 3 (- ln tan (\frac{x}{2}) + 2 ln tan (\frac{x}{2}) + 1) + C

What is the integral of (3cos(x))/(sin^2(x)+sin(x)) ?
The integral of (3cos(x))/(sin^2(x)+sin(x)) is -3(-ln|tan(x/2)|+2ln|tan(x/2)+1|)+C