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

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

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

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

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

Solution steps

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

Apply u-substitution

: \int \frac{2 u}{- u ^{2} + 1} d u

= \int \frac{2 u}{- u ^{2} + 1} d u

Take the constant out:

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

= 2 \cdot \int \frac{u}{- u ^{2} + 1} d u

Apply u-substitution

: \int - \frac{1}{2 v} d v

= 2 \cdot \int - \frac{1}{2 v} d v

Take the constant out:

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

= 2 (- \frac{1}{2} \cdot \int \frac{1}{v} d v)

Use the common integral:

\int \frac{1}{v} d v = ln (∣ v ∣)

= 2 (- \frac{1}{2} ln ∣ v ∣)

Substitute back

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

Simplify

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

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

Add a constant to the solution

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

4 x^{2} - 4, x^{8} - 1

\frac{\partial}{\partial x} (y + 3 x z + x^{3})

f (x) = x^{3} - 4 x^{2} - 9 x + 11

\int_{- \infty}^{\infty} x^{2} e^{\frac{- x ^{2}}{2}} d x

\frac{d y}{d x}, x y + 4 e^{y} = 4 e

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