What is the integral of e^{2x}sin(3e^{2x}+1) ?

The integral of e^{2x}sin(3e^{2x}+1) is -1/6 cos(3e^{2x}+1)+C

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integral of e^{2x}sin(3e^{2x}+1)

\int e^{2 x} sin (3 e^{2 x} + 1) d x

- \frac{1}{6} cos (3 e^{2 x} + 1) + C

Solution steps

\int e^{2 x} sin (3 e^{2 x} + 1) d x

Apply u-substitution

: \int \frac{sin ( u )}{6} d u

= \int \frac{sin ( u )}{6} d u

Take the constant out:

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

= \frac{1}{6} \cdot \int sin (u) d u

Use the common integral:

\int sin (u) d u = - cos (u)

= \frac{1}{6} (- cos (u))

Substitute back

u = 3 e^{2 x} + 1

= \frac{1}{6} (- cos (3 e^{2 x} + 1))

Simplify

= - \frac{1}{6} cos (3 e^{2 x} + 1)

Add a constant to the solution

= - \frac{1}{6} cos (3 e^{2 x} + 1) + C

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tan (x^{2})

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What is the integral of e^{2x}sin(3e^{2x}+1) ?
The integral of e^{2x}sin(3e^{2x}+1) is -1/6 cos(3e^{2x}+1)+C