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

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

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

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

\frac{1}{16} (ln \frac{sin ( 4 x )}{2} + 1 - ln \frac{sin ( 4 x )}{2} - 1) + C

Solution steps

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

Rewrite using trig identities

= \int \frac{cos ( 4 x )}{cos ^{2} ( 4 x ) + 3} d x

Apply u-substitution

= \int \frac{cos ( u )}{4 ( cos ^{2} ( u ) + 3 )} d u

Take the constant out:

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

= \frac{1}{4} \cdot \int \frac{cos ( u )}{cos ^{2} ( u ) + 3} d u

Rewrite using trig identities

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

Apply u-substitution

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

Apply Integral Substitution

= \frac{1}{4} \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}{4} \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 = \frac{ln ∣ w + 1 ∣}{2} - \frac{ln ∣ w - 1 ∣}{2}

= \frac{1}{4} \cdot \frac{1}{2} (\frac{ln ∣ w + 1 ∣}{2} - \frac{ln ∣ w - 1 ∣}{2})

Substitute back

= \frac{1}{4} \cdot \frac{1}{2} \frac{ln \frac{s i n ( 4 x )}{2} + 1}{2} - \frac{ln \frac{s i n ( 4 x )}{2} - 1}{2}

Simplify

\frac{1}{4} \cdot \frac{1}{2} \frac{ln \frac{s i n ( 4 x )}{2} + 1}{2} - \frac{ln \frac{s i n ( 4 x )}{2} - 1}{2} : \frac{1}{16} (ln \frac{sin ( 4 x )}{2} + 1 - ln \frac{sin ( 4 x )}{2} - 1)

= \frac{1}{16} (ln \frac{sin ( 4 x )}{2} + 1 - ln \frac{sin ( 4 x )}{2} - 1)

Add a constant to the solution

= \frac{1}{16} (ln \frac{sin ( 4 x )}{2} + 1 - ln \frac{sin ( 4 x )}{2} - 1) + C

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