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

The integral of 1/(1+sin^2(x)) is 1/(sqrt(2))arctan(sqrt(2)tan(x))+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of 1/(1+sin^2(x))

integral

\frac{1}{1 + sin ^{2} ( x )}

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

Solution steps

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

Apply Trigonometric Substitution

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

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

Apply Integral Substitution

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

Take the constant out:

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

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

Use the common integral:

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

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

Substitute back

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

Add a constant to the solution

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

t^{2} e^{t}

5 y^{2} (e^{x})^{2}

- 0.5 x^{2}

sin (x) + 2 cos (x)

sin^{3} (2 x)

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