What is the integral of sin(2θ)e^{sin^2(θ)} ?

The integral of sin(2θ)e^{sin^2(θ)} is e^{sin^2(θ)}+C

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integral of sin(2θ)e^{sin^2(θ)}

\int sin (2 θ) e^{s i n^{2} (θ)} d θ

e^{s i n^{2} (θ)} + C

Solution steps

\int sin (2 θ) e^{s i n^{2} (θ)} d θ

Rewrite using trig identities

= \int sin (2 θ) e^{\frac{1 - c o s ( 2 θ )}{2}} d θ

Apply u-substitution

: \int \frac{e ^{\frac{u}{2}}}{2} d u

= \int \frac{e ^{\frac{u}{2}}}{2} 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 e^{\frac{u}{2}} d u

Apply u-substitution

: \int e^{v} \cdot 2 d v

= \frac{1}{2} \cdot \int e^{v} \cdot 2 d v

Take the constant out:

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

= \frac{1}{2} \cdot 2 \cdot \int e^{v} d v

Use the common integral:

\int e^{v} d v = e^{v}

= \frac{1}{2} \cdot 2 e^{v}

Substitute back

= \frac{1}{2} \cdot 2 e^{\frac{1 - c o s ( 2 θ )}{2}}

Simplify

\frac{1}{2} \cdot 2 e^{\frac{1 - c o s ( 2 θ )}{2}} : e^{s i n^{2} (θ)}

= e^{s i n^{2} (θ)}

Add a constant to the solution

= e^{s i n^{2} (θ)} + C

\frac{\partial}{\partial x} (I n ∣ x - y ∣)

x \to 2 lim (e^{\frac{3}{( 2 - x )}})

\frac{5 x ^{3}}{2 x - 2}

y^{^{''}} - y^{^{'}} - 12 y = 2 e^{3 x}

\frac{\partial}{\partial v} (u (v) (v)^{2} e^{- v})

What is the integral of sin(2θ)e^{sin^2(θ)} ?
The integral of sin(2θ)e^{sin^2(θ)} is e^{sin^2(θ)}+C