how do u integrate sinxcosx?!?!?

The derivative of sine is cosine, which is found outside of the function. So use u-substitution:∫ sin(x)·cos(x) dxLet u = sin(x)Then du = cos(x) dx→ ∫ u du = u²/2 + CReverse substitution:= ½ sin²(x) + C— — — — — — — — — — — — — — — — — — — — — — — — — — — — —You can actually do the same with cosine: ∫ sin(x)·cos(x) dxLet u = cos(x)Then du = -sin(x) dx→ ∫ -u du= -u²/2 + C= -cos²(x) + C,Note that though these seem different, they vary by only a constant. Using the pythagorean identity:sin²θ + cos²θ = 1cos²θ = 1 – sin²θ→ -[ 1 – sin²(x) ] + C,= -1 + sin²(x) + C,Negative one plus a constant is just another constant:= sin²(x) + C,,And this is identical to the one above.– – – –Note: the method below skips u-substitution, which of course can be done, but it’s just a matter of writing the process more easily.… and the one below that uses a double angle identity.Actually, since there should be a constant added, that can become:[ 1 – cos(2·x) ] / 4 + C,= 1/4 – cos(2·x)/4 + C,= -cos(2·x)/4 + CBut the double angle formula isn’t needed. All of these antiderivatives (and more, I’d imagine) are within a constant of each other and equally valid.... Show More

Integral Sinxcosx... Show More

Integrate Sinx Cosx... Show More

https://shorturl.im/axuYO sinx cosx = (1/2) sin 2x => Integral = (1/2) * (- 1/2 ) * cos 2x + c = – (1/4) cos2x + c.... Show More

This Site Might Help You.RE:how do u integrate sinxcosx?!?!?... Show More

∫ sin x cos x dx = ∫ sin x d [ sin x]= sin^2 x / 2 + CThis is without using a substitution.... Show More

Int{sinx cosx dx} since d(sinx) = cosx = Int{sinx d(sinx) } bet. = (sinx)^2 / 2 + const.= (1 – cos 2x) / 4 also.... Show More