Question

Hi! I'm a newbie in integral calculus, can I get some explanation on how to integrate this one: ∫cos1/3xsin1/3xdx Thank you!

Answer 1

Can you write it in a more clear way?

Answer 2

I hope this is alright.

Answer 3

do you know the formula for \(\sin 2\theta =...?\)

Answer 4

oh, and i see you are new here, so Hi,
\(\Huge \mathcal{\text{Welcome To OpenStudy}\ddot\smile} \)

Answer 5

Oh, thanks! I'm not sure maybe sin2θ=2sinθcosθ? I'm not really sure

Answer 6

yes, thats correct! :)
so what can you write
\(\large \sin(1/3 \: x)\cos (1/3 \: x)=...?\)
as ?

Answer 7

hint : \(\large \sin(1/3 \: x)\cos (1/3 \: x)=(1/2)\large [2\sin(1/3 \: x)\cos (1/3 \: x)]=...?\)

Answer 8

I'm sorry for asking, but where did the 1/2 came from (base on your hint). Sorry, I'm really lost here.

Answer 9

sinθcosθ= sin2θ/2
right ?

Answer 10

so,
sin (1/3 x) cos (1/3 x) = sin (2/3 x) /2
did you get this ??

Answer 11

$$\int\cos\left(\frac13x\right)\sin\left(\frac13x\right)\mathrm{d}x=\frac12\int2\cos\left(\frac13x\right)\sin\left(\frac13x\right)\mathrm{d}x$$We know that \(2\sin\theta\cos\theta=\sin2\theta\), so:$$\frac12\int2\cos\left(\frac13x\right)\sin\left(\frac13x\right)\mathrm{d}x=\frac12\int\sin\left(\frac23x\right)\mathrm{d}x=\frac12\times\frac32\int\sin u\,\mathrm{d}u\text{ where }u=\frac23x$$I trust you can finish it off from here.

Answer 12

You could also use a substitution from the start. Let \(u=\sin\left(\frac13x\right)\). It follows that \(\mathrm{d}u=\frac13\cos\left(\frac13x\right)\mathrm{d}x\), so \(3\,\mathrm{d}u=\cos\left(\frac13x\right)\mathrm{d}x\).

Answer 13

Thank you for guiding me in solving this problem. I'll try my best to finish it correctly. Thank you again.