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

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

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

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

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

Solution steps

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

Apply u-substitution

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

= \int - \frac{1}{16 + u ^{2}} d u

Take the constant out:

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

= - \int \frac{1}{16 + u ^{2}} d u

Apply Integral Substitution

= - \int \frac{1}{4 ( v ^{2} + 1 )} d v

Take the constant out:

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

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

Use the common integral:

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

= - \frac{1}{4} arctan (v)

Substitute back

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

Add a constant to the solution

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

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