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

The integral of 2^x*sin(x) is -(2^xcos(x))/(1+ln^2(2))+(ln(2)*2^xsin(x))/(1+ln^2(2))+C

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integral of 2^x*sin(x)

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

- \frac{2 ^{x} cos ( x )}{1 + ln ^{2} ( 2 )} + \frac{ln ( 2 ) \cdot 2 ^{x} sin ( x )}{1 + ln ^{2} ( 2 )} + C

Solution steps

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

Apply Integration By Parts

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

Take the constant out:

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

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

Apply Integration By Parts

= - 2^{x} cos (x) - (- ln (2) (2^{x} sin (x) - \int 2^{x} ln (2) sin (x) d x))

Take the constant out:

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

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

Therefore

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

Isolate

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

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

Add a constant to the solution

= - \frac{2 ^{x} cos ( x )}{1 + ln ^{2} ( 2 )} + \frac{ln ( 2 ) \cdot 2 ^{x} sin ( x )}{1 + ln ^{2} ( 2 )} + C

What is the integral of 2^x*sin(x) ?
The integral of 2^x*sin(x) is -(2^xcos(x))/(1+ln^2(2))+(ln(2)*2^xsin(x))/(1+ln^2(2))+C