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

The integral of (sin(x)cos(x))/(cos^2(x)+4) is -1/2 ln|cos(2x)+9|+C

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integral of (sin(x)cos(x))/(cos^2(x)+4)

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

- \frac{1}{2} ln ∣ cos (2 x) + 9 ∣ + C

Solution steps

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

Rewrite using trig identities

= \int \frac{sin ( 2 x )}{2 ( 4 + cos ^{2} ( x ) )} d x

Take the constant out:

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

= \frac{1}{2} \cdot \int \frac{sin ( 2 x )}{4 + cos ^{2} ( x )} d x

Rewrite using trig identities

= \frac{1}{2} \cdot \int \frac{2 sin ( 2 x )}{cos ( 2 x ) + 9} d x

Take the constant out:

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

= \frac{1}{2} \cdot 2 \cdot \int \frac{sin ( 2 x )}{cos ( 2 x ) + 9} d x

Apply u-substitution

= \frac{1}{2} \cdot 2 \cdot \int - \frac{1}{2 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 2 (- \frac{1}{2} \cdot \int \frac{1}{u} d u)

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= \frac{1}{2} \cdot 2 (- \frac{1}{2} ln ∣ u ∣)

Substitute back

u = cos (2 x) + 9

= \frac{1}{2} \cdot 2 (- \frac{1}{2} ln ∣ cos (2 x) + 9 ∣)

Simplify

\frac{1}{2} \cdot 2 (- \frac{1}{2} ln ∣ cos (2 x) + 9 ∣) : - \frac{1}{2} ln ∣ cos (2 x) + 9 ∣

= - \frac{1}{2} ln ∣ cos (2 x) + 9 ∣

Add a constant to the solution

= - \frac{1}{2} ln ∣ cos (2 x) + 9 ∣ + C

What is the integral of (sin(x)cos(x))/(cos^2(x)+4) ?
The integral of (sin(x)cos(x))/(cos^2(x)+4) is -1/2 ln|cos(2x)+9|+C