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

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

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

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

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

Solution steps

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

Simplify

\frac{sin ( arctan ( x ) )}{1 + x ^{2}} : \frac{x}{( 1 + x ^{2} ) 1 + x ^{2}}

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

Apply u-substitution

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

= \int \frac{1}{2 u u} 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{1}{u u} d u

Simplify

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

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

Apply the Power Rule

: - \frac{2}{u}

= \frac{1}{2} (- \frac{2}{u})

Substitute back

u = 1 + x^{2}

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

Simplify

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

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

Add a constant to the solution

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

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