What is the integral of e^{2θ} ?

The integral of e^{2θ} is 1/2 e^{2θ}+C

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integral of e^{2θ}

\int e^{2 θ} d θ

\frac{1}{2} e^{2 θ} + C

Solution steps

\int e^{2 θ} d θ

Apply u-substitution

: \int e^{u} \frac{1}{2} d u

= \int e^{u} \frac{1}{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^{u} d u

Use the common integral:

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

= \frac{1}{2} e^{u}

Substitute back

u = 2 θ

= \frac{1}{2} e^{2 θ}

Add a constant to the solution

= \frac{1}{2} e^{2 θ} + C

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

\frac{d}{d x} (2 a cos (2 x) - 2 b sin (2 x))

r \frac{2 x}{3 x ^{2} - 4}

f (x) = \frac{x ^{2} - 2 x}{x}

(x^{2} - 2 x)^{3}