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

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

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Answer 1 · 2018-05-17T23:34:10

$\int 5^{x} sin x d x = \frac{5^{x} (ln 5 sin x - cos x)}{1 + {ln}^{2} 5} + C$ $int 5^x sinx dx = (5^x(ln5sinx- cosx))/(1+ln^2 5)+C$

Explanation:

Integrate by parts:

$\int 5^{x} sin x d x = \int 5^{x} \frac{d}{d x} (- cos x) d x$ $int 5^x sinx dx = int 5^x d/dx (-cosx) dx$

$\int 5^{x} sin x d x = = - 5^{x} cos x + \int cos x \frac{d}{d x} (5^{x}) d x$ $int 5^x sinx dx = = -5^xcosx + int cosx d/dx(5^x) dx$

$\int 5^{x} sin x d x = = - 5^{x} cos x + ln 5 \int cos x 5^{x} d x$ $int 5^x sinx dx = = -5^xcosx + ln5 int cosx 5^xdx$

and then again:

$\int 5^{x} sin x d x = = - 5^{x} cos x + ln 5 \int \frac{d}{d x} (sin x) 5^{x} d x$ $int 5^x sinx dx = = -5^xcosx + ln5 int d/dx(sinx) 5^xdx$

$\int 5^{x} sin x d x = = - 5^{x} cos x + ln 5 sin x 5^{x} - ln 5 \int sin x \frac{d}{d x} (5^{x}) d x$ $int 5^x sinx dx = = -5^xcosx + ln5sinx5^x - ln5 int sinx d/dx( 5^x)dx$

$\int 5^{x} sin x d x = = 5^{x} (ln 5 sin x - cos x) - {ln}^{2} 5 \int sin x 5^{x} d x$ $int 5^x sinx dx = = 5^x(ln5sinx- cosx) - ln^2 5 int sinx5^xdx$

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

$(1 + {ln}^{2} 5) \int 5^{x} sin x d x = 5^{x} (ln 5 sin x - cos x) + C$ $(1+ln^2 5)int 5^x sinx dx = 5^x(ln5sinx- cosx)+C$

$\int 5^{x} sin x d x = \frac{5^{x} (ln 5 sin x - cos x)}{1 + {ln}^{2} 5} + C$ $int 5^x sinx dx = (5^x(ln5sinx- cosx))/(1+ln^2 5)+C$

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

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Explanation:

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