How do I integrate Cos x Sin x dx?

you can solve in different ways you will get diff answer but they will be right

1) put sinx = t

cosxdx = dt

∫ tdt = t^2/2 +c = 1/2sin^2 X +c

2) sinxcosx = 1/2 sin2x

∫ 1/2sin2x dx = -1/4 cos2x +c

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There are different ways to solve this, but here's the way I'd go about it: ∫ sin(x) / cos^3 (x) dx = ∫ sin(x) [ cos (x) ]^-3 dx = (-1/2) ∫ sin(x) * -2[ cos (x) ]^-3 dx = (1/2) ∫ -sin(x) * -2[ cos (x) ]^-3 dx = (1/2) cos^-2 (x) + c If you use "u substitition" by letting u = cos(x) so that du = -sin(x) dx, etc. you'll get the right answer. I just noticed that you have cos(x) raised to some power, and that the derivative of cosine is -sin(x), so I did the "chain rule" in reverse. Somebody else answered (1/2)tan²(x) + c, which is not wrong. If you use the fact that sin^2 (x) + cos^2(x) = 1, and apply it to the other answer, then you have: (1/2) (1/cos^2 (x)) + c = (1/2) [sin^2 (x) + cos^2(x)]/cos^2 (x)) + c = (1/2) [tan^2 (x) + 1] + c = (1/2)tan^2 (x) + 1/2 + c = (1/2)tan^2 (x) + (1/2 + c) = (1/2)tan^2 (x) + (a constant)

Integrate Sinxcosx

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How do I integrate Cos x Sin x dx?

How do I integrate Cos x Sin x dx? I'm working on an engineering problem and having trouble getting up to speed with simple integral calculus. Any help is much appreciated!

Integral Of Cosxsinx

There is a rule states => sinA cosB =( sin(A+B)+sin(A-B) ) /2

integral sinx cosx =integral (sin(2x)+sin(0))/2

= ( -cos(2x) /2 ) +0

= ( -cos(2x) /4 ) +c

^^

;)

How do I integrate Cos^4(3x) sin^4(3x) ?

how to integrate cos x sin nx ?

cosx sinx dx = 1/2sin^2 x +c ,meaning half of sinx square

also cosxsinx = -1/2cos^2 x + c ,means half of cos square negative.

∫cosx sinx dx

= ∫sinx dsinx

= (1/2)sin^2(x) + c