What is the integral of sin^2(e^x) ?

The integral of sin^2(e^x) is 1/2 (x-\Ci(2e^x))+C

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integral of sin^2(e^x)

\int sin^{2} (e^{x}) d x

\frac{1}{2} (x - Ci (2 e^{x})) + C

Solution steps

\int sin^{2} (e^{x}) d x

Apply u-substitution

: \int \frac{sin ^{2} ( u )}{u} d u

= \int \frac{sin ^{2} ( u )}{u} d u

Rewrite using trig identities

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

Take the constant out:

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

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

Expand

\frac{1 - cos ( 2 u )}{u} : \frac{1}{u} - \frac{cos ( 2 u )}{u}

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

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

\int \frac{1}{u} d u = ln ∣ u ∣

\int \frac{cos ( 2 u )}{u} d u = Ci (2 u)

= \frac{1}{2} (ln ∣ u ∣ - Ci (2 u))

Substitute back

u = e^{x}

= \frac{1}{2} (ln ∣ e^{x} ∣ - Ci (2 e^{x}))

ln ∣ e^{x} ∣ = x

= \frac{1}{2} (x - Ci (2 e^{x}))

Add a constant to the solution

= \frac{1}{2} (x - Ci (2 e^{x})) + C

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What is the integral of sin^2(e^x) ?
The integral of sin^2(e^x) is 1/2 (x-\Ci(2e^x))+C