how do u integrate sinxcosx?!?!?

Question

How do you integrate sin(x)cos(x)?

Anonymous · Accepted Answer

The derivative of sine is cosine, which is found outside of the function. So use u-substitution:

∫ sin(x)·cos(x) dx

Let u = sin(x)

Then du = cos(x) dx

→ ∫ u du

= u²/2 + C

Reverse substitution:

= ½ sin²(x) + C

— — — — — — — — — — — — — — — — — — — — — — — — — — — — —

You can actually do the same with cosine:

∫ sin(x)·cos(x) dx

Let 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 = 1

cos²θ = 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 + C

But 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.

Sydney Bernier · Answer

Integral Sinxcosx

Carol Donnelly · Answer

Integrate Sinx Cosx

Anonymous · Answer

https://shorturl.im/axuYOsinx cosx = (1/2) sin 2x => Integral = (1/2) * (- 1/2 ) * cos 2x + c = &ndash; (1/4) cos2x + c.

Moses Ledner IV · Answer

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RE:
how do u integrate sinxcosx?!?!?

Anonymous · Answer

∫ sin x cos x dx = ∫ sin x d [ sin x]
= sin^2 x / 2 + C
This is without using a substitution.

Anonymous · Answer

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

Anonymous

how do u integrate sinxcosx?!?!?

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