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

The integral of 2^x*sin(3x) is -(3*2^xcos(3x))/(9+ln^2(2))+(ln(2)*2^xsin(3x))/(9+ln^2(2))+C

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integral of 2^x*sin(3x)

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

- \frac{3 \cdot 2 ^{x} cos ( 3 x )}{9 + ln ^{2} ( 2 )} + \frac{ln ( 2 ) \cdot 2 ^{x} sin ( 3 x )}{9 + ln ^{2} ( 2 )} + C

Solution steps

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

Apply Integration By Parts

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

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

Take the constant out:

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

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

Apply Integration By Parts

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

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

Take the constant out:

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

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

Therefore

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

Isolate

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

= - \frac{3 \cdot 2 ^{x} cos ( 3 x )}{9 + ln ^{2} ( 2 )} + \frac{ln ( 2 ) \cdot 2 ^{x} sin ( 3 x )}{9 + ln ^{2} ( 2 )}

Add a constant to the solution

= - \frac{3 \cdot 2 ^{x} cos ( 3 x )}{9 + ln ^{2} ( 2 )} + \frac{ln ( 2 ) \cdot 2 ^{x} sin ( 3 x )}{9 + ln ^{2} ( 2 )} + C

n = 1 \sum \infty \frac{6}{n}

n = 1 \sum \infty \frac{1}{n ^{\frac{7}{2}}}

y^{^{''}} + 28 y = 0

\int \frac{y ^{2}}{8} d y

x sin (\frac{3}{x})

What is the integral of 2^x*sin(3x) ?
The integral of 2^x*sin(3x) is -(3*2^xcos(3x))/(9+ln^2(2))+(ln(2)*2^xsin(3x))/(9+ln^2(2))+C