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

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

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

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

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

Solution steps

\int \frac{2 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

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

Rewrite using trig identities

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

Apply u-substitution

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

= 2 \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

= 2 (- \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}

= 2 (- \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

= 2 (- (- \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 ∣

= 2 (- (- ln ∣ u ∣ + 2 ln ∣ u + 1 ∣))

Substitute back

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

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

Simplify

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

Add a constant to the solution

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

x \to 1 lim (\frac{x}{x ^{2} + 2})

f (x) = \frac{3 π x ^{2}}{2}

\int \frac{1 + x}{1 - x ^{2}} d x

\frac{2}{3 x + 1}

\frac{\partial}{\partial y} (2 - y - x^{2})

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