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

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

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

x^{2} - x - 2

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

x y^{3}

\frac{d y}{d x}, y = x + \frac{1}{x}

\frac{d}{d x} (\frac{3 x ^{2} - 1}{2 x + 5})

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