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 April 10th, 2011, 10:05 PM #1 Senior Member   Joined: Jan 2009 Posts: 344 Thanks: 3 How would you go about solving these integrals? $\text{\int{e^{2x}sin(x)}dx}$ $\text{\int{e^{-x}cos(2x)}dx}$
 April 10th, 2011, 10:18 PM #2 Global Moderator     Joined: Nov 2009 From: Northwest Arkansas Posts: 2,766 Thanks: 4 Re: How would you go about solving these integrals? http://en.wikipedia.org/wiki/Integration_by_parts Note: this isn't just a "hey goofball, look it up on wikipedia" answer. They work through a similar example in the section "powers of e"... After 2(?) iterations of IBP, your original integral will appear on BOTH sides of the equation (with different coefficients). Combine like terms and solve for your integral.
 April 10th, 2011, 10:27 PM #3 Senior Member     Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,209 Thanks: 517 Math Focus: Calculus/ODEs Re: How would you go about solving these integrals? 1.) $\int e^{2x}\sin x\,dx$ Using integration by parts, let: $u=\sin x\:\therefore\:du=\cos x\,dx$ $dv=e^{2x}\,dx\:\therefore\:v=\frac{1}{2}e^{2x}$ Thus, we have: $\int e^{2x}\sin x\,dx=\frac{1}{2}e^{2x}\sin x-\frac{1}{2}\int e^{2x}\cos x\,dx$ Now, let's isolate: $\int e^{2x}\cos x\,dx$ Let: $u=\cos x\:\therefore\:du=-\sin x\,dx$ $dv=e^{2x}\,dx\:\therefore\:v=\frac{1}{2}e^{2x}$ Thus, we have: $\int e^{2x}\cos x\,dx=\frac{1}{2}e^{2x}\cos x+\frac{1}{2}\int e^{2x}\sin x\,dx$ Substituting for this into the previous result, we have: $\int e^{2x}\sin x\,dx=\frac{1}{2}e^{2x}\sin x-\frac{1}{2}$$\frac{1}{2}e^{2x}\cos x+\frac{1}{2}\int e^{2x}\sin x\,dx$$$ $\int e^{2x}\sin x\,dx=\frac{1}{2}e^{2x}\sin x-\frac{1}{4}e^{2x}\cos x-\frac{1}{4}\int e^{2x}\sin x\,dx$ $\frac{5}{4}\int e^{2x}\sin x\,dx=\frac{1}{2}e^{2x}\sin x-\frac{1}{4}e^{2x}\cos x$ $\int e^{2x}\sin x\,dx=\frac{4}{5}$$\frac{1}{2}e^{2x}\sin x-\frac{1}{4}e^{2x}\cos x$$+C$ $\int e^{2x}\sin x\,dx=\frac{e^{2x}}{5}$$2\sin x-\cos x$$+C$ 2.) $\int e^{-x}\cos(2x)\,dx$ Using integration by parts, let: $u=\cos(2x)\:\therefore\:du=-2\sin(2x)\,dx$ $dv=e^{-x}\,dx\:\therefore\:v=-e^{-x}$ Thus, we have: $\int e^{-x}\cos(2x)\,dx=-e^{-x}\cos(2x)-2\int e^{-x}\sin(2x)\,dx$ Now, let's isolate: $\int e^{-x}\sin(2x)\,dx$ Using integration by parts, let: $u=\sin(2x)\:\therefore\:du=2\cos(2x)\,dx$ $dv=e^{-x}\,dx\:\therefore\:v=-e^{-x}$ Thus, we have: $\int e^{-x}\sin(2x)\,dx=-e^{-x}\sin(2x)+2\int e^{-x}\cos(2x)\,dx$ Substituting this into the previous step, we have: $\int e^{-x}\cos(2x)\,dx=-e^{-x}\cos(2x)-2$$-e^{-x}\sin(2x)+2\int e^{-x}\cos(2x)\,dx$$$ $\int e^{-x}\cos(2x)\,dx=-e^{-x}\cos(2x)+2e^{-x}\sin(2x)-4\int e^{-x}\cos(2x)\,dx$ $5\int e^{-x}\cos(2x)\,dx=-e^{-x}\cos(2x)+2e^{-x}\sin(2x)=e^{-x}$$2\sin(2x)-\cos(2x)$$$ $\int e^{-x}\cos(2x)\,dx=\frac{e^{-x}}{5}$$2\sin(2x)-\cos(2x)$$+C$
 April 10th, 2011, 11:12 PM #4 Senior Member     Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,209 Thanks: 517 Math Focus: Calculus/ODEs Re: How would you go about solving these integrals? In general, we may set up the following integral: $\int e^{ax}\sin$$bx+c$$\,dx$ Using integration by parts, let: $u=\sin$$bx+c$$\:\therefore\:du=b\cos$$bx+c$$\,dx$ $dv=e^{ax}\,dx\:\therefore\:v=\frac{1}{a}e^{ax}$ Thus, we have: $\int e^{ax}\sin$$bx+c$$\,dx=\frac{1}{a}e^{ax}\sin$$bx+c$$-\frac{b}{a}\int e^{ax}\cos$$bx+c$$\,dx$ Now, let's isolate: $\int e^{ax}\cos$$bx+c$$\,dx$ Using integration by parts, let: $u=\cos$$bx+c$$\:\therefore\:du=-b\sin$$bx+c$$\,dx$ $dv=e^{ax}\,dx\:\therefore\:v=\frac{1}{a}e^{ax}$ Thus, we have: $\int e^{ax}b\cos$$bx+c$$\,dx=\frac{1}{a}e^{ax}\cos$$bx+ c$$+\frac{b}{a}\int e^{ax}b\sin$$bx+c$$\,dx$ Substituting into the previous step, we have: $\int e^{ax}\sin$$bx+c$$\,dx=\frac{1}{a}e^{ax}\sin$$bx+c$$-\frac{b}{a}$$\frac{1}{a}e^{ax}\cos\(bx+c$$+\frac{b }{a}\int e^{ax}b\sin$$bx+c$$\,dx\)$ $\int e^{ax}\sin$$bx+c$$\,dx=\frac{1}{a}e^{ax}\sin$$bx+c$$-\frac{b}{a^2}e^{ax}\cos$$bx+c$$-\frac{b^2}{a^2}\int e^{ax}b\sin$$bx+c$$\,dx$ $\frac{a^2+b^2}{a^2}\int e^{ax}\sin$$bx+c$$\,dx=\frac{1}{a}e^{ax}\sin$$bx+c$$-\frac{b}{a^2}e^{ax}\cos$$bx+c$$$ $\frac{a^2+b^2}{a}\int e^{ax}\sin$$bx+c$$\,dx=e^{ax}$$\sin\(bx+c$$-\frac{b}{a}\cos$$bx+c$$\)+C$ $\int e^{ax}\sin$$bx+c$$\,dx=\frac{ae^{ax}}{a^2+b^2}$$\s in\(bx+c$$-\frac{b}{a}\cos$$bx+c$$\)+C$ $\int e^{ax}\sin$$bx+c$$\,dx=\frac{e^{ax}}{a^2+b^2}$$a\s in\(bx+c$$-b\cos$$bx+c$$\)+C$ So, for problem 1, we have a = 2, b = 1, c = 0, so the result would be: $\frac{e^{2x}}{5}$$2\sin\(x$$-\cos$$x$$\)+C$ For problem 2, we have a = -1, b = -2, c = ?/2, so the result would be: $\frac{e^{-x}}{5}$$-\sin\(-2x+\frac{\pi}{2}$$+2\cos$$-2x+\frac{\pi}{2}$$\)+C=\frac{e^{-x}}{5}$$2\sin\(2x$$-\cos$$2x$$\)+C$
 April 11th, 2011, 09:37 AM #5 Senior Member   Joined: May 2008 From: York, UK Posts: 1,300 Thanks: 0 Re: How would you go about solving these integrals? Generally speaking, I find the easiest way to do integrals involving combinations of exponentials and trig functions is to simply use the identities $\cos x=\frac12$$e^{ix}+e^{-ix}$$,\ \ \sin x = \frac1{2i}$$e^{ix}-e^{-ix}$$,$ multiply out and then integrate the resulting sum of exponential functions. You can then use the identities to simplify it back into trig functions where possible.
 April 11th, 2011, 06:30 PM #6 Senior Member   Joined: Apr 2007 Posts: 2,140 Thanks: 0 Apply integration by parts, which should be easy for the given two problems.
 April 11th, 2011, 06:48 PM #7 Global Moderator     Joined: Nov 2009 From: Northwest Arkansas Posts: 2,766 Thanks: 4 Re: How would you go about solving these integrals? All in favor of integration by parts, say "ay"!

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