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I have a problem with solving one integral due to lack of experience in the integration logic. So i have line integral of kind 2 as follows:

$$\int_0^{2 \pi}(2x) \ dx + (3yx) \ dy,$$

given by $C: x = 4\cos(2t), y = 3\sin(2t)$ I intentionally give the example because maybe there is more easy, smart way to do this problem. So i evaluate it and i get $$\int_0^{2 \pi}-64\cos(2t)\sin(2t) + 216\cos^2(2t)\sin(2t)$$ so after some trig identities i have came to this integral $$ \int_{0}^{2 \pi}-64\sin(2t)\cos(2t)+108\sin(4t)\cos(2t)$$ so i decided to make two integrals from this one and i have a problem with integrating $I_2$ $$I_1= -64\int_0^{2 \pi}\sin(2t)\cos(2t) \\ I_2= 108\int_0^{2 \pi}\sin(4t)\cos(2t)$$ Here is my actual question i am usually using $u$ substitution here and i get $$u = \cos(2t) , \ du=-2\sin(2t) \\dt = \frac{du}{2\cos(t)}$$ but this is valid when i have the same angles i want to ask is is still right to do it when i have $\sin(4t) \ ,\cos(2t)$ and if it is how is should be done or is there another way to integrate this part. Thank you in any help in advance.

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    $\begingroup$ suggestion: put $\backslash$ before trigonometric functions. It looks nicer. Also, Knuth intended it to be this way. $\endgroup$ – uniquesolution Jun 17 at 11:51
  • $\begingroup$ The integration limits are weird. You say $\int_0^{2\pi} 2x{\rm d}x$ as if $x$ is to be integrated between $0$ and $2\pi$, but then you say $x = 4\cos(2t)$ so it's clearly $t$ that is the integration variable (and $x$ is limited to $[-4,4]$ so cannot be integrated over $[0,2\pi]$). $\endgroup$ – Winther Jun 17 at 12:02
  • $\begingroup$ @Winther yes this are the parameters for the line integral you substitute x and y with the parameters given by $C$ its over a curved line. I dunno what is possible the problem is given by with that bounds $t \in [0,2 \pi]$ $\endgroup$ – Boris Borovski Jun 17 at 12:04
  • $\begingroup$ @BorisBorovski Check that I didn't mess anything up in my editing. $\endgroup$ – The Pointer Jun 17 at 12:04
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    $\begingroup$ @The Pointer Everything is in place. Thank you for the edit. $\endgroup$ – Boris Borovski Jun 17 at 12:05
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Hint: $$\sin(a) \cos(b) = \frac{\sin(a+b)}{2} + \frac{\sin(a-b)}{2}$$

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Yes, there's another (simpler) way of computing these integrals: linearisation of these trigonometric polynomials: $$\sin 2t\cos 2t=\tfrac12\sin 4t,\qquad \sin 4t\cos 2t=\tfrac12\bigl(\sin(4t+2t)+\sin(4t-2t)\bigr).$$

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  • $\begingroup$ @Robert Israel Thanks for the help but in the end is its not legal to make substitution with different angles ? $\endgroup$ – Boris Borovski Jun 17 at 12:15

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