What is the integral of (-sin(x))/(1+cos^2(x)) ?

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

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

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

Solution steps

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

Take the constant out:

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

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

Rewrite using trig identities

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

Apply u-substitution

: \int \frac{2 u}{u ^{4} + 1} d u

= - \int \frac{2 u}{u ^{4} + 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 ^{4} + 1} d u

Apply u-substitution

: \int \frac{1}{2 ( v ^{2} + 1 )} d v

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

Take the constant out:

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

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

Use the common integral:

\int \frac{1}{v ^{2} + 1} d v = arctan (v)

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

Substitute back

= - 2 \cdot \frac{1}{2} arctan (tan^{2} (\frac{x}{2}))

Simplify

- 2 \cdot \frac{1}{2} arctan (tan^{2} (\frac{x}{2})) : - arctan (tan^{2} (\frac{x}{2}))

= - arctan (tan^{2} (\frac{x}{2}))

Add a constant to the solution

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

\int (\frac{100 + x ^{2}}{10 x}) d x

\int x^{π - 1} d x

\int \frac{x ^{2}}{x ^{6} + 2} d x

\int \frac{x ^{3} + x + 1}{( x + 1 ) ( x ^{2} + 4 ) ^{2}} d x

\int \frac{2 x}{x ^{2} - 81} d x

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