What is the integral of (sin(arctan(2x)))/(1+4x^2) ?

The integral of (sin(arctan(2x)))/(1+4x^2) is -1/(2sqrt(1+4x^2))+C

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integral of (sin(arctan(2x)))/(1+4x^2)

\int \frac{sin ( arctan ( 2 x ) )}{1 + 4 x ^{2}} d x

- \frac{1}{2 1 + 4 x ^{2}} + C

Solution steps

\int \frac{sin ( arctan ( 2 x ) )}{1 + 4 x ^{2}} d x

Apply u-substitution

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

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

Apply u-substitution

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

= \frac{1}{2} \cdot \int \frac{1}{2 v v} d v

Take the constant out:

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

= \frac{1}{2} \cdot \frac{1}{2} \cdot \int \frac{1}{v v} d v

Simplify

v v : v^{\frac{3}{2}}

= \frac{1}{2} \cdot \frac{1}{2} \cdot \int \frac{1}{v ^{\frac{3}{2}}} d v

Apply the Power Rule

: - \frac{2}{v}

= \frac{1}{2} \cdot \frac{1}{2} (- \frac{2}{v})

Substitute back

= \frac{1}{2} \cdot \frac{1}{2} - \frac{2}{1 + ( 2 x ) ^{2}}

Simplify

\frac{1}{2} \cdot \frac{1}{2} - \frac{2}{1 + ( 2 x ) ^{2}} : - \frac{1}{2 1 + 4 x ^{2}}

= - \frac{1}{2 1 + 4 x ^{2}}

Add a constant to the solution

= - \frac{1}{2 1 + 4 x ^{2}} + C

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