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

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

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

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

- ln 1 + cos^{2} (x) + C

Solution steps

\int \frac{2 sin ( x ) cos ( 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

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

Rewrite using trig identities

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

Take the constant out:

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

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

Apply u-substitution

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

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

Take the constant out:

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

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

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= 2 \cdot \frac{1}{2} (- ln ∣ u ∣)

Substitute back

u = 1 + cos^{2} (x)

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

Simplify

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

= - ln 1 + cos^{2} (x)

Add a constant to the solution

= - ln 1 + cos^{2} (x) + C

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