What is the integral of (sin(5x))/(16+cos^2(5x)) ?

The integral of (sin(5x))/(16+cos^2(5x)) is -1/20 arctan(1/4 cos(5x))+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of (sin(5x))/(16+cos^2(5x))

\int \frac{sin ( 5 x )}{16 + cos ^{2} ( 5 x )} d x

- \frac{1}{20} arctan (\frac{1}{4} cos (5 x)) + C

Solution steps

\int \frac{sin ( 5 x )}{16 + cos ^{2} ( 5 x )} d x

Apply u-substitution

: \int \frac{sin ( u )}{5 ( 16 + cos ^{2} ( u ) )} d u

= \int \frac{sin ( u )}{5 ( 16 + cos ^{2} ( u ) )} d u

Take the constant out:

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

= \frac{1}{5} \cdot \int \frac{sin ( u )}{16 + cos ^{2} ( u )} d u

Apply u-substitution

: \int - \frac{1}{16 + v ^{2}} d v

= \frac{1}{5} \cdot \int - \frac{1}{16 + v ^{2}} d v

Take the constant out:

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

= \frac{1}{5} (- \int \frac{1}{16 + v ^{2}} d v)

Apply Integral Substitution

= \frac{1}{5} (- \int \frac{1}{4 ( 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}{5} (- \frac{1}{4} \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}{5} (- \frac{1}{4} arctan (w))

Substitute back

= \frac{1}{5} (- \frac{1}{4} arctan (\frac{cos ( 5 x )}{4}))

Simplify

\frac{1}{5} (- \frac{1}{4} arctan (\frac{cos ( 5 x )}{4})) : - \frac{1}{20} arctan (\frac{1}{4} cos (5 x))

= - \frac{1}{20} arctan (\frac{1}{4} cos (5 x))

Add a constant to the solution

= - \frac{1}{20} arctan (\frac{1}{4} cos (5 x)) + C

\frac{\partial}{\partial x} (6 x + 5 y)

\int \frac{24 t + 6 t ^{3}}{8 t ^{2} + t ^{4}} d t

e^{- 2 t - 3}

\frac{d}{d x} (\frac{x}{π})

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

What is the integral of (sin(5x))/(16+cos^2(5x)) ?
The integral of (sin(5x))/(16+cos^2(5x)) is -1/20 arctan(1/4 cos(5x))+C