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VCE Stuff => VCE Mathematics => VCE Mathematics/Science/Technology => VCE Subjects + Help => VCE Specialist Mathematics => Topic started by: M-D on June 07, 2013, 10:50:55 am

Title: integration using trig identities
Post by: M-D on June 07, 2013, 10:50:55 am

hi,

could someone please integrate these:

1) $\int\2sin^4(4x)cos(2x) dx$

2) $\int\cos^5(2x) dx$

thanks

Title: Re: integration using trig identities
Post by: bobbyz0r on June 07, 2013, 12:13:43 pm

1. $(1/3)sin^3(2x) - (1/5)sin^5(2x) + c$

Title: Re: integration using trig identities
Post by: M-D on June 07, 2013, 01:37:15 pm

i am very sorry. in (1) I had actually made a mistake in the question while putting it in Latex form.
here is the question:

$\int2\sin^4(2x)cos(2x) dx$

this is how i went about it:

$2\int\sin^4(2x)cos(2x) dx$

$\int(1-\cos^2(2x))^2cos(2x) dx$

let $u= cos(2x)$

$\frac{du}{dx}=-2sin(2x)$

$\frac{du}{-2sin(2x)}=dx$

what should now be done?

thanks

Title: Re: integration using trig identities
Post by: e^1 on June 07, 2013, 01:40:25 pm

Quote from: M-D on June 07, 2013, 10:50:55 am

hi,

could someone please integrate these:

1) $\int\2sin^4(4x)cos(2x) dx$

2) $\int\cos^5(2x) dx$

thanks

2. Hint: Use the Pythagorean trigonometric identity.

Spoiler

$\begin{alignedat}{1} \int {\cos^5(2x)}dx = \int{\cos(2x)(1-\sin^2(2x))^2} dx \\\\ \text{Let } u = \sin(2x) \\ \frac{du}{dx} = 2\cos(2x) \\ \therefore dx = \frac{du}{2\cos(2x)} \\\\ \int{\cos(x)(1-\sin^2(x))^2} dx &= \frac{1}{2}\int{(1-u^2)^2}du \\ &= \frac{1}{2}\int{1 - 2u^2 + u^4}du \\ &= \frac{1}{2}u - \frac{1}{3}u^3 + \frac{1}{10}u^5 +C \\ &= \frac{1}{2}\sin(2x) - \frac{1}{3}\sin^3(2x) + \frac{1}{10}\sin^5(2x) +C \end{alignedat}$

Title: Re: integration using trig identities
Post by: b^3 on June 07, 2013, 01:42:53 pm

1. You need to pick a $u$ that will allow you to simplify the integral into something that we can work with. So in this case the easiest way about going about the question is to pick a $u$ that will allow you to get rid of the $\cos(2x)$ . So you don't need to change the $\sin^{4}(2x)$ , but how can we pick a $u$ so that the derivative will be something like $\cos(2x)$ ? We can pick $u=\sin(2x)$ .

Try with that and see how you go.

EDIT:

Spoiler

$\begin{alignedat}{1}\int2\sin^{4}\left(2x\right)\cos\left(2x\right)dx \\ \text{Let }u=\sin\left(2x\right) \\ \frac{du}{dx}=2\cos\left(2x\right) \\ \cos\left(2x\right)dx=\frac{1}{2}du \\ \int2\sin^{4}\left(2x\right)\cos\left(2x\right)dx & =2\int\frac{1}{2}\times u^{4}du \\ & =\int u^{4}du \\ & =\frac{1}{5}u^{5}+C \\ & =\frac{1}{5}\sin^{5}\left(2x\right)+C \end{alignedat} $

Title: Re: integration using trig identities
Post by: M-D on June 07, 2013, 02:39:31 pm

thank you very much. I also need help with another question which goes:

$\int\sin^4(x)cos^5(x) dx$

thanks

Title: Re: integration using trig identities
Post by: b^3 on June 07, 2013, 02:44:31 pm

When you have and odd power and an even power of sines and cosines, you can expand the term that has the odd power so that you have the product of even powers and one odd power of the trig function. The even powers then can be turned into the other trig function (so if it's cosine, into sine, and if it's sine into cosine) by using the Pythagorean trigonometric identity. Then you can make a simple $u$ substitution on the trig functions that were from the even powers to get rid of that odd power term, expand the $u$ 's and then integrate simply.

E.g. for the Above you could express it as $\begin{alignedat}{1}\int\sin^{4}\left(x\right)\cos^{2}\left(x\right)\cos^{2}\left(x\right)\cos\left(x\right)dx & =\int\sin^{4}\left(x\right)\left(1-\sin^{2}\left(x\right)\right)^{2}\cos\left(x\right)dx\end{alignedat}$

Then what do you think our $u$ substitution will be to make it simpler?

Hope that helps :)

EDIT:

Try before you check this

$\begin{alignedat}{1}\int\sin^{4}\left(x\right)\cos^{5}\left(x\right)dx & =\int\sin^{4}\left(x\right)\left(1-\sin^{2}\left(x\right)\right)^{2}\cos\left(x\right)dx \\ \text{Let }u=\sin\left(x\right) \\ \frac{du}{dx}=\cos\left(x\right) \\ \cos\left(x\right)dx=du \\ \int\sin^{4}\left(x\right)\cos^{5}\left(x\right)dx & =\int u^{4}\left(1-u^{2}\right)^{2}du \\ & =\int u^{4}\left(1-2u^{2}+u^{4}\right)du \\ & =\int\left(u^{4}-2u^{6}+u^{8}\right)du \\ & =\frac{1}{5}u^{5}-\frac{2}{7}u^{7}+\frac{1}{9}u^{9}+C \\ & =\frac{1}{5}\sin^{5}\left(x\right)-\frac{2}{7}\sin^{7}\left(x\right)+\frac{1}{9}\sin^{9}\left(x\right)+C \end{alignedat} $

Title: Re: integration using trig identities
Post by: M-D on June 07, 2013, 03:54:46 pm

thanks b^3 and everyone else that helped. :)

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VCE Stuff => VCE Mathematics => VCE Mathematics/Science/Technology => VCE Subjects + Help => VCE Specialist Mathematics => Topic started by: M-D on June 07, 2013, 10:50:55 am