What is the integral of+(cos(3x))/(1+sin^2(3x)) ?

The integral of+(cos(3x))/(1+sin^2(3x)) is 1/3 arctan(sin(3x))+C

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integral of+(cos(3x))/(1+sin^2(3x))

\int + \frac{cos ( 3 x )}{1 + sin ^{2} ( 3 x )} d x

\frac{1}{3} arctan (sin (3 x)) + C

Solution steps

\int \frac{cos ( 3 x )}{1 + sin ^{2} ( 3 x )} d x

Apply u-substitution

: \int \frac{cos ( u )}{3 ( 1 + sin ^{2} ( u ) )} d u

= \int \frac{cos ( u )}{3 ( 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}{3} \cdot \int \frac{cos ( u )}{1 + sin ^{2} ( u )} d u

Apply u-substitution

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

= \frac{1}{3} \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}{3} arctan (v)

Substitute back

= \frac{1}{3} arctan (sin (3 x))

Add a constant to the solution

= \frac{1}{3} arctan (sin (3 x)) + C

\frac{d}{d x} (sin (3 x - 2))

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

\int \frac{62 x ^{9}}{2835} d x

x \to 1 + lim (\frac{1}{x ^{2} - 4 x + 3})

\frac{\partial}{\partial x} (tan (x^{3} + y))

What is the integral of+(cos(3x))/(1+sin^2(3x)) ?
The integral of+(cos(3x))/(1+sin^2(3x)) is 1/3 arctan(sin(3x))+C