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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

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

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

Solution steps

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

Simplify

= \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

\int_{- 1}^{1} 5 d x

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

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

\frac{d}{d x} (sin (x - 1) x^{3})

\frac{\partial}{\partial z} (\frac{x y}{x ^{2} + y ^{2}})

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