What is the integral of ((sin(31x))/(1+cos^2(31x))) ?

The integral of ((sin(31x))/(1+cos^2(31x))) is 1/31 arctan(tan^2((31x)/2))+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

integral of ((sin(31x))/(1+cos^2(31x)))

\int (\frac{sin ( 31 x )}{1 + cos ^{2} ( 31 x )}) d x

\frac{1}{31} arctan (tan^{2} (\frac{31 x}{2})) + C

Solution steps

\int \frac{sin ( 31 x )}{1 + cos ^{2} ( 31 x )} d x

Apply u-substitution

= \int \frac{sin ( u )}{31 ( 1 + 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}{31} \cdot \int \frac{sin ( u )}{1 + cos ^{2} ( u )} d u

Rewrite using trig identities

= \frac{1}{31} \cdot \int \frac{sin ( u )}{- sin ^{2} ( u ) + 2} d u

Apply u-substitution

= \frac{1}{31} \cdot \int \frac{2 v}{v ^{4} + 1} d v

Take the constant out:

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

= \frac{1}{31} \cdot 2 \cdot \int \frac{v}{v ^{4} + 1} d v

Apply u-substitution

= \frac{1}{31} \cdot 2 \cdot \int \frac{1}{2 ( 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}{31} \cdot 2 \cdot \frac{1}{2} \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}{31} \cdot 2 \cdot \frac{1}{2} arctan (w)

Substitute back

= \frac{1}{31} \cdot 2 \cdot \frac{1}{2} arctan (tan^{2} (\frac{31 x}{2}))

Simplify

\frac{1}{31} \cdot 2 \cdot \frac{1}{2} arctan (tan^{2} (\frac{31 x}{2})) : \frac{1}{31} arctan (tan^{2} (\frac{31 x}{2}))

= \frac{1}{31} arctan (tan^{2} (\frac{31 x}{2}))

Add a constant to the solution

= \frac{1}{31} arctan (tan^{2} (\frac{31 x}{2})) + C

d er i v a t i v e e^{t + 7}

x \to - 7 + lim (\frac{- 6 x}{x + 7})

d er i v a t i v e f (θ) = 3 θ

\frac{d}{d x} (\frac{8 x y z}{( x ^{2} + 1 ) ^{2}})

\int x^{3} - 3 x^{2} - x + 3 d x

What is the integral of ((sin(31x))/(1+cos^2(31x))) ?
The integral of ((sin(31x))/(1+cos^2(31x))) is 1/31 arctan(tan^2((31x)/2))+C