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

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

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

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

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

Solution steps

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

Apply Trigonometric Substitution

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

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

Apply Integral Substitution

= \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

= \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)

= \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

\frac{\partial}{\partial x} (2 1 - y^{2})

(14 + x) \frac{d y}{d x} = 5 y

\frac{d y}{d x} = \frac{( x - 1 )}{y - 1}

\frac{d v}{d t} - \frac{4 v}{t} = 8 t^{3}

\int 1 - y d y

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