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

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

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

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

4 arctan (tan^{2} (\frac{x}{2})) + C

Solution steps

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

Take the constant out:

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

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

Rewrite using trig identities

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

Apply u-substitution

: \int \frac{2 u}{u ^{4} + 1} d u

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

Take the constant out:

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

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

Apply u-substitution

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

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

Take the constant out:

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

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

Use the common integral:

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

= 4 \cdot 2 \cdot \frac{1}{2} arctan (v)

Substitute back

= 4 \cdot 2 \cdot \frac{1}{2} arctan (tan^{2} (\frac{x}{2}))

Simplify

4 \cdot 2 \cdot \frac{1}{2} arctan (tan^{2} (\frac{x}{2})) : 4 arctan (tan^{2} (\frac{x}{2}))

= 4 arctan (tan^{2} (\frac{x}{2}))

Add a constant to the solution

= 4 arctan (tan^{2} (\frac{x}{2})) + C

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