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

The integral of 1/(a^2sin^2(x)+b^2cos^2(x)) is 1/(ab)arctan(a/b tan(x))+C

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integral of 1/(a^2sin^2(x)+b^2cos^2(x))

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

\frac{1}{ab} arctan (\frac{a}{b} tan (x)) + C

Solution steps

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

Apply Trigonometric Substitution

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

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

Apply Integral Substitution

= \int \frac{1}{ab ( 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}{ab} \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}{ab} arctan (v)

Substitute back

= \frac{1}{ab} arctan (\frac{a}{b} tan (x))

Add a constant to the solution

= \frac{1}{ab} arctan (\frac{a}{b} tan (x)) + C

\frac{\partial}{\partial x} (x e^{- x} (y^{2} - 4 y))

\frac{\partial}{\partial x} (\frac{1}{x ^{2} + a ^{2}})

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

\frac{7}{s ^{2} + 5 s + 10}

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