What is the integral of 3^xsin(2x) ?

The integral of 3^xsin(2x) is -(2*3^xcos(2x))/(4+ln^2(3))+(ln(3)*3^xsin(2x))/(4+ln^2(3))+C

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integral of 3^xsin(2x)

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

- \frac{2 \cdot 3 ^{x} cos ( 2 x )}{4 + ln ^{2} ( 3 )} + \frac{ln ( 3 ) \cdot 3 ^{x} sin ( 2 x )}{4 + ln ^{2} ( 3 )} + C

Solution steps

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

Apply Integration By Parts

: - \frac{1}{2} \cdot 3^{x} cos (2 x) - \int - \frac{1}{2} ln (3) \cdot 3^{x} cos (2 x) d x

= - \frac{1}{2} \cdot 3^{x} cos (2 x) - \int - \frac{1}{2} ln (3) \cdot 3^{x} cos (2 x) d x

Take the constant out:

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

= - \frac{1}{2} \cdot 3^{x} cos (2 x) - (- \frac{1}{2} ln (3) \cdot \int 3^{x} cos (2 x) d x)

Apply Integration By Parts

: 3^{x} \cdot \frac{1}{2} sin (2 x) - \int 3^{x} ln (3) \frac{1}{2} sin (2 x) d x

= - \frac{1}{2} \cdot 3^{x} cos (2 x) - (- \frac{1}{2} ln (3) (3^{x} \cdot \frac{1}{2} sin (2 x) - \int 3^{x} ln (3) \frac{1}{2} sin (2 x) d x))

Take the constant out:

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

= - \frac{1}{2} \cdot 3^{x} cos (2 x) - (- \frac{1}{2} ln (3) (3^{x} \cdot \frac{1}{2} sin (2 x) - ln (3) \frac{1}{2} \cdot \int 3^{x} sin (2 x) d x))

Therefore

\int 3^{x} sin (2 x) d x = - \frac{1}{2} \cdot 3^{x} cos (2 x) - (- \frac{1}{2} ln (3) (3^{x} \cdot \frac{1}{2} sin (2 x) - ln (3) \frac{1}{2} \cdot \int 3^{x} sin (2 x) d x))

Isolate

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

= - \frac{2 \cdot 3 ^{x} cos ( 2 x )}{4 + ln ^{2} ( 3 )} + \frac{ln ( 3 ) \cdot 3 ^{x} sin ( 2 x )}{4 + ln ^{2} ( 3 )}

Add a constant to the solution

= - \frac{2 \cdot 3 ^{x} cos ( 2 x )}{4 + ln ^{2} ( 3 )} + \frac{ln ( 3 ) \cdot 3 ^{x} sin ( 2 x )}{4 + ln ^{2} ( 3 )} + C

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What is the integral of 3^xsin(2x) ?
The integral of 3^xsin(2x) is -(2*3^xcos(2x))/(4+ln^2(3))+(ln(3)*3^xsin(2x))/(4+ln^2(3))+C