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

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

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

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

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

Solution steps

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

Rewrite using trig identities

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

Apply u-substitution

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

= \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

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

Apply u-substitution

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

= 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

= 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)

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

Substitute back

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

Simplify

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

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

Add a constant to the solution

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

\int sin (3 x) sin (2 x) sin (x) d x

\frac{1}{s + \frac{1}{a}}

x, \frac{1}{x ^{2}}, 2

\frac{d}{d a} (sin^{4} (a) - cos^{4} (a))

\frac{1}{4} y^{^{''}} + 4 y = 2 tan (4 t) - \frac{1}{3} e^{3 t}

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