How do you find the integral of $(sinx)(5^x) dx$ ?

Question

3653 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-05-17T23:34:10

$int 5^x sinx dx = (5^x(ln5sinx- cosx))/(1+ln^2 5)+C$

Integrate by parts:

$int 5^x sinx dx = int 5^x d/dx (-cosx) dx$

$int 5^x sinx dx = = -5^xcosx + int cosx d/dx(5^x) dx$

$int 5^x sinx dx = = -5^xcosx + ln5 int cosx 5^xdx$

and then again:

$int 5^x sinx dx = = -5^xcosx + ln5 int d/dx(sinx) 5^xdx$

$int 5^x sinx dx = = -5^xcosx + ln5sinx5^x - ln5 int sinx d/dx( 5^x)dx$

$int 5^x sinx dx = = 5^x(ln5sinx- cosx) - ln^2 5 int sinx5^xdx$

The integral now appears on both sides of the equation and we can solve for it:

$(1+ln^2 5)int 5^x sinx dx = 5^x(ln5sinx- cosx)+C$

$int 5^x sinx dx = (5^x(ln5sinx- cosx))/(1+ln^2 5)+C$

How do you find the integral of (sinx)(5^x) dx (sinx)(5^x) dx?