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calc
find integral using table of integrals ) integral sin^4xdx this the formula i used integral sin^n xdx =1/n sin^n1xcosx +n1/n integral sin^n2 using the formula this is what i got: integral sin^4xdx=1/4sin^3xcosx+3/4 integral
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integral from 0 to 2pi of isin(t)e^(it)dt. I know my answer should be pi. **I pull i out because it is a constant. My work: let u=e^(it) du=ie^(it)dt dv=sin(t) v=cos(t) i integral sin(t)e^(it)dt= e^(it)cos(t)+i*integral
asked by Ashley on April 16, 2015 
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That's the same as the integral of sin^2 x dx. Use integration by parts. Let sin x = u and sin x dx = dv v = cos x du = cos x dx The integral is u v  integral of v du = sinx cosx + integral of cos^2 dx which can be rewritten
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s integral endpoints are 0 and pi/2 i need to find the integral of sin^2 (2x) dx. i know that the answer is pi/4, but im not sure how to get to it. i know: s sin^2(2x)dx= 1/2 [1cos (4x)] dx, but then i'm confused. The indefinite
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Question : Integrate [x/(1+(sin a*sin x))] from 0 to pi My first thought was to apply integrate f(x) dx= f(ax) dx method Which simplified the integral into; 2I = integrate [pi/(1+(sin a*sin x))] dx , cancelling out x Then I made
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I'm trying to find the convolution f*g where f(t)=g(t)=sin(t). I set up the integral and proceed to do integration by parts twice, but it keeps working out to 0=0 or sin(t)=sin(t). How am I supposed to approach it? integral
asked by chuck on December 3, 2009 
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i'm having trouble evaluating the integral at pi/2 and 0. i know: s (at pi/2 and 0) sin^2 (2x)dx= s 1/2[1cos(2x)]dx= s 1/2(xsin(4x))dx= (x/2) 1/8[sin (4x)] i don't understand how you get pi/4 You made a few mistakes, check
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How would I integrate the following by parts: Integral of: (x^2)(sin (ax))dx, where a is any constant. Just like you did x^2 exp(x) below. Also partial integration is not the easiest way to do this integral. You can also use this
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