What is the integral of e^{2q}cos(3e^q) ?

The integral of e^{2q}cos(3e^q) is 1/9 (3e^qsin(3e^q)+cos(3e^q))+C

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integral of e^{2q}cos(3e^q)

\int e^{2 q} cos (3 e^{q}) d q

\frac{1}{9} (3 e^{q} sin (3 e^{q}) + cos (3 e^{q})) + C

Solution steps

\int e^{2 q} cos (3 e^{q}) d q

Apply u-substitution

: \int u cos (3 u) d u

= \int u cos (3 u) d u

Apply u-substitution

: \int \frac{v cos ( v )}{9} d v

= \int \frac{v cos ( v )}{9} d v

Take the constant out:

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

= \frac{1}{9} \cdot \int v cos (v) d v

Apply Integration By Parts

: v sin (v) - \int sin (v) d v

= \frac{1}{9} (v sin (v) - \int sin (v) d v)

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

= \frac{1}{9} (v sin (v) - (- cos (v)))

Substitute back

= \frac{1}{9} (3 e^{q} sin (3 e^{q}) - (- cos (3 e^{q})))

Simplify

= \frac{1}{9} (3 e^{q} sin (3 e^{q}) + cos (3 e^{q}))

Add a constant to the solution

= \frac{1}{9} (3 e^{q} sin (3 e^{q}) + cos (3 e^{q})) + C

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What is the integral of e^{2q}cos(3e^q) ?
The integral of e^{2q}cos(3e^q) is 1/9 (3e^qsin(3e^q)+cos(3e^q))+C