integral of x sinx dx Explanation Use integration by parts which takes the form udv uv vdu For udv x2sin x dx we let u x2 du dx 2x du 2xdx dv sin x dx dv sin x dx v cos x Thus substituting these into the integration by parts formula we see that x2sin x dx x2cos x 2xcos x dx
Int 0 oo sinx x dx pi 2 We seek I int 0 oo sinx x dx Let g x sinx x g x sin x x sinx x Thus g x is an even function and as such 2I int oo oo sinx x dx Consider the complex based function f z e iz z Which has a simple pole at z 0 we then consider the contour integral J oint C f z dz oint C e iz z dz where z in CC Where Sep 12 2015 Integrate by parts twice using u e2x both times Explanation After the second integration by parts you ll have e2xsinxdx e2x cosx 2e2xsinx 4 e2xsinxdx Note that the last integral is the same as the one we want Call it I for now I e2x cosx 2e2xsinx 4I So I 1 5 e2xcosx 2e2xsinx C
integral of x sinx dx
integral of x sinx dx
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Q42 1 1 sinx Dx Integral Of 1 1 sinx Dx Indefinte Integration
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Therefore x sinx 1 cos x dx x tan x 2 C where C is an arbitrary constant Suggest Corrections 60 How do you find the integral of ex sin x How do you find the integral of e sin x Calculus Techniques of Integration Integration by Parts
Then plugging into the IBP formula gives us sinx e2x dx sinx 1 2 e2x 1 2 e2x cosx dx e2x sinx dx 1 2 e2xsinx 1 2 I Inserting this result into A we get I 1 2e2xcosx 1 2 1 2 e2xsinx 1 2I A I 1 2e2x cosx 1 4e2xsinx 1 4I A 4I 2e2x cosx e2xsinx I 4A 5I 2e2xcosx Explanation I x 1 sinx dx x 1 sinx 1 sin2x dx x 1 sinx cos2x dx I xsec2xdx xsecxtanxdx Integration by parts we get I xtanx 1 tanxdx x secx 1 secxdx I xtanx ln secx xsecx ln secx tanx c
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