What is the integral of (sin(19x))/(1+cos^2(19x)) ?

The integral of (sin(19x))/(1+cos^2(19x)) is 1/19 arctan(tan^2((19x)/2))+C

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integral of (sin(19x))/(1+cos^2(19x))

\int \frac{sin ( 19 x )}{1 + cos ^{2} ( 19 x )} d x

\frac{1}{19} arctan (tan^{2} (\frac{19 x}{2})) + C

Solution steps

\int \frac{sin ( 19 x )}{1 + cos ^{2} ( 19 x )} d x

Apply u-substitution

: \int \frac{sin ( u )}{19 ( 1 + cos ^{2} ( u ) )} d u

= \int \frac{sin ( u )}{19 ( 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}{19} \cdot \int \frac{sin ( u )}{1 + cos ^{2} ( u )} d u

Rewrite using trig identities

= \frac{1}{19} \cdot \int \frac{sin ( u )}{- sin ^{2} ( u ) + 2} d u

Apply u-substitution

: \int \frac{2 v}{v ^{4} + 1} d v

= \frac{1}{19} \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}{19} \cdot 2 \cdot \int \frac{v}{v ^{4} + 1} d v

Apply u-substitution

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

= \frac{1}{19} \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}{19} \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}{19} \cdot 2 \cdot \frac{1}{2} arctan (w)

Substitute back

= \frac{1}{19} \cdot 2 \cdot \frac{1}{2} arctan (tan^{2} (\frac{19 x}{2}))

Simplify

\frac{1}{19} \cdot 2 \cdot \frac{1}{2} arctan (tan^{2} (\frac{19 x}{2})) : \frac{1}{19} arctan (tan^{2} (\frac{19 x}{2}))

= \frac{1}{19} arctan (tan^{2} (\frac{19 x}{2}))

Add a constant to the solution

= \frac{1}{19} arctan (tan^{2} (\frac{19 x}{2})) + C

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What is the integral of (sin(19x))/(1+cos^2(19x)) ?
The integral of (sin(19x))/(1+cos^2(19x)) is 1/19 arctan(tan^2((19x)/2))+C