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

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

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

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

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

Solution steps

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

Rewrite using trig identities

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

Apply u-substitution

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

= \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

= - \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}

= - \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

= - (- \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 ∣

= - (- ln ∣ u ∣ + 2 ln ∣ u + 1 ∣)

Substitute back

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

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

Simplify

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

Add a constant to the solution

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

y^{^{''}} + 34 y = 0

d x + d y e^{6 x} = 0

f (x) = e^{1}

\int 4 - 4 x^{2} d x

∣ x - 9 ∣, \frac{x}{2}

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