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

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

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

\int \frac{sin ( 2 x )}{1 + e ^{c o s^{2} (x)}} d x

- ln (e^{c o s^{2} (x)}) + ln e^{c o s^{2} (x)} + 1 + C

Solution steps

\int \frac{sin ( 2 x )}{1 + e ^{c o s^{2} (x)}} d x

Apply u-substitution

: \int - \frac{1}{u ( 1 + u )} d u

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

Take the constant out:

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

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

Take the partial fraction of

\frac{1}{u ( 1 + u )} : \frac{1}{u} - \frac{1}{u + 1}

= - \int \frac{1}{u} - \frac{1}{u + 1} d u

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

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

\int \frac{1}{u} d u = ln ∣ u ∣

\int \frac{1}{u + 1} d u = ln ∣ u + 1 ∣

= - (ln ∣ u ∣ - ln ∣ u + 1 ∣)

Substitute back

u = e^{c o s^{2} (x)}

= - (ln e^{c o s^{2} (x)} - ln e^{c o s^{2} (x)} + 1)

Simplify

- (ln e^{c o s^{2} (x)} - ln e^{c o s^{2} (x)} + 1) : - ln (e^{c o s^{2} (x)}) + ln e^{c o s^{2} (x)} + 1

= - ln (e^{c o s^{2} (x)}) + ln e^{c o s^{2} (x)} + 1

Add a constant to the solution

= - ln (e^{c o s^{2} (x)}) + ln e^{c o s^{2} (x)} + 1 + C

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