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

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

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

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

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

Solution steps

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

Apply Trigonometric Substitution

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

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

Take the constant out:

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

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

Apply Integral Substitution

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

\frac{v ^{2}}{v ^{2} + 1} = - \frac{1}{v ^{2} + 1} + 1

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

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

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

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

\int 1 d v = v

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

Substitute back

= - \frac{1}{2 2} (- arctan (2 cot (x)) + 2 cot (x))

Simplify

- \frac{1}{2 2} (- arctan (2 cot (x)) + 2 cot (x)) : \frac{1}{2 2} arctan (2 cot (x)) - \frac{1}{2} cot (x)

= \frac{1}{2 2} arctan (2 cot (x)) - \frac{1}{2} cot (x)

Add a constant to the solution

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

y = - x, y = 2 - x, y = 0

y = \frac{3 x ^{2}}{2 x + 4}

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

\frac{\partial}{\partial x} (2 A y (L - x))

\int sec^{2} (7 - 4 x) d x

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