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

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

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

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

ln 1 + sin^{2} (x) + C

Solution steps

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

Rewrite using trig identities

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

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= 2 \cdot \frac{1}{2} ln ∣ u ∣

Substitute back

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

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

Simplify

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

= ln 1 + sin^{2} (x)

Add a constant to the solution

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

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

y^{^{''}} - y^{^{'}} - 56 y = 0

\frac{3 x - 1}{e ^{x}}

\frac{\partial}{\partial x} (1 - \frac{y}{x})

x (cos (x^{2}))

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