What is the integral of (sin(2x))/(cos^2(2x)+5) ?

The integral of (sin(2x))/(cos^2(2x)+5) is -1/(2sqrt(5))arctan(1/(sqrt(5))cos(2x))+C

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integral of (sin(2x))/(cos^2(2x)+5)

\int \frac{sin ( 2 x )}{cos ^{2} ( 2 x ) + 5} d x

- \frac{1}{2 5} arctan (\frac{1}{5} cos (2 x)) + C

Solution steps

\int \frac{sin ( 2 x )}{cos ^{2} ( 2 x ) + 5} d x

Apply u-substitution

: \int \frac{sin ( u )}{2 ( cos ^{2} ( u ) + 5 )} d u

= \int \frac{sin ( u )}{2 ( cos ^{2} ( u ) + 5 )} d u

Take the constant out:

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

= \frac{1}{2} \cdot \int \frac{sin ( u )}{cos ^{2} ( u ) + 5} d u

Apply u-substitution

: \int - \frac{1}{v ^{2} + 5} d v

= \frac{1}{2} \cdot \int - \frac{1}{v ^{2} + 5} d v

Take the constant out:

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

= \frac{1}{2} (- \int \frac{1}{v ^{2} + 5} d v)

Apply Integral Substitution

= \frac{1}{2} (- \int \frac{1}{5 ( w ^{2} + 1 )} d w)

Take the constant out:

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

= \frac{1}{2} (- \frac{1}{5} \cdot \int \frac{1}{w ^{2} + 1} d w)

Use the common integral:

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

= \frac{1}{2} (- \frac{1}{5} arctan (w))

Substitute back

= \frac{1}{2} (- \frac{1}{5} arctan (\frac{cos ( 2 x )}{5}))

Simplify

\frac{1}{2} (- \frac{1}{5} arctan (\frac{cos ( 2 x )}{5})) : - \frac{1}{2 5} arctan (\frac{1}{5} cos (2 x))

= - \frac{1}{2 5} arctan (\frac{1}{5} cos (2 x))

Add a constant to the solution

= - \frac{1}{2 5} arctan (\frac{1}{5} cos (2 x)) + C

x \to 1 lim (4 x)

\int sec^{2} (t) d t

x \to 0 + lim (\frac{( e ^{x} )}{2 ^{x}})

\int cos (x) (4 + 5 sin^{2} (x)) d x

\frac{d}{d θ} (sin (2 θ))

What is the integral of (sin(2x))/(cos^2(2x)+5) ?
The integral of (sin(2x))/(cos^2(2x)+5) is -1/(2sqrt(5))arctan(1/(sqrt(5))cos(2x))+C