What is the integral of (cos(6x))/(1+sin^2(6x)) ?

The integral of (cos(6x))/(1+sin^2(6x)) is 1/6 arctan(sin(6x))+C

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integral of (cos(6x))/(1+sin^2(6x))

\int \frac{cos ( 6 x )}{1 + sin ^{2} ( 6 x )} d x

\frac{1}{6} arctan (sin (6 x)) + C

Solution steps

\int \frac{cos ( 6 x )}{1 + sin ^{2} ( 6 x )} d x

Apply u-substitution

: \int \frac{cos ( u )}{6 ( 1 + sin ^{2} ( u ) )} d u

= \int \frac{cos ( u )}{6 ( 1 + sin ^{2} ( u ) )} d u

Take the constant out:

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

= \frac{1}{6} \cdot \int \frac{cos ( u )}{1 + sin ^{2} ( u )} d u

Apply u-substitution

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

= \frac{1}{6} \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}{6} arctan (v)

Substitute back

= \frac{1}{6} arctan (sin (6 x))

Add a constant to the solution

= \frac{1}{6} arctan (sin (6 x)) + C

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What is the integral of (cos(6x))/(1+sin^2(6x)) ?
The integral of (cos(6x))/(1+sin^2(6x)) is 1/6 arctan(sin(6x))+C