2007-12-26 18:51:01 · 12 answers · asked by mathstudent 2 in Science & Mathematics ➔ Mathematics
Let y = sin x
dy = cos x dx
∫ sin x cos x dx
= ∫ y dy
= (y^2)/2 + c
= (sin^2 x)/2 + c
2007-12-26 18:59:52 · answer #1 · answered by Blake 3 · 12⤊ 2⤋
Integral Of Sinxcosx
2016-12-16 12:51:13 · answer #2 · answered by finnen 4 · 0⤊ 0⤋
This Site Might Help You.
RE:
How do you find the integral of sin(x)cos(x)?
2015-08-10 04:49:22 · answer #3 · answered by Anonymous · 0⤊ 0⤋
Integral Of Sinx Cosx
2016-10-01 06:03:14 · answer #4 · answered by ? 4 · 0⤊ 0⤋
The other answers are correct. Here's an alternative solution, whose answer is different in form but equivalent to the others.
Note first that sin(2x)=2sin(x)cos(x). Hence,
Integral sin(x)cos(x) dx = Integral [sin(2x)]/2 dx
= -cos(2x)/4 + C
2007-12-26 19:17:52 · answer #5 · answered by Chuck 3 · 9⤊ 0⤋
How do you find the integral of sin(x)cos(y)?
2014-12-29 15:03:55 · answer #6 · answered by Anonymous · 0⤊ 0⤋
I = ∫ sin x cos x dx
I = (1/2) ∫ 2 sin x cos x dx
I = (1/2) ∫ sin 2x dx
I = (- 1/4 ) cos 2x + C
2007-12-26 20:54:29 · answer #7 · answered by Como 7 · 7⤊ 0⤋
LET SINX= t derivative of Sinx=cosx Cosx=dt%dx integer of t T ka square by 2 put t=sinx ka square by 2 answer
2016-07-21 03:19:40 · answer #8 · answered by kalu 1 · 0⤊ 0⤋
Find the equation at which the derivative is sin(x)cos(x).
(1/2)Sin^2 (x) + c
2007-12-26 19:02:17 · answer #9 · answered by JONATHAN L 1 · 2⤊ 3⤋
For the best answers, search on this site https://shorturl.im/aw6PR
Integral of SinXCosX(SinX+CosX) = =[Integral of [SinXCosX(SinX) dx ] ]+ [ Integral of [SinXCosX(CosX) dx ] ] =[Integral of{[SinX] ^2] d[SinX]} ] - [Integral of{[CosX] ^2] d[CosX]} ] =[ [[SinX]^3]/3 ] - [[CosX] ^3]/3] +constant
2016-04-05 09:47:52 · answer #10 · answered by ? 4 · 0⤊ 0⤋