What is the integral of (cos(7x))/(1+sin^2(7x)) ?

The integral of (cos(7x))/(1+sin^2(7x)) is 1/7 arctan(sin(7x))+C

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integral of (cos(7x))/(1+sin^2(7x))

\int \frac{cos ( 7 x )}{1 + sin ^{2} ( 7 x )} d x

\frac{1}{7} arctan (sin (7 x)) + C

Solution steps

\int \frac{cos ( 7 x )}{1 + sin ^{2} ( 7 x )} d x

Apply u-substitution

: \int \frac{cos ( u )}{7 ( 1 + sin ^{2} ( u ) )} d u

= \int \frac{cos ( u )}{7 ( 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}{7} \cdot \int \frac{cos ( u )}{1 + sin ^{2} ( u )} d u

Apply u-substitution

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

= \frac{1}{7} \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}{7} arctan (v)

Substitute back

= \frac{1}{7} arctan (sin (7 x))

Add a constant to the solution

= \frac{1}{7} arctan (sin (7 x)) + C

f (x) = x^{2} (x - 2)^{4}

\frac{d y}{d x} = \frac{y}{x ^{2} + 1}

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

\frac{d}{d x} (\frac{- 2 x - y}{x + 2 y})

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

What is the integral of (cos(7x))/(1+sin^2(7x)) ?
The integral of (cos(7x))/(1+sin^2(7x)) is 1/7 arctan(sin(7x))+C