Definite Integration of Trig Functions

[donnac10]

It's me again.. >_< I need another help.

Evaluate the definite integral.

sin²(9x)cos²(9x) dx​

​

*from 6 to 13

I tried doing substitution.. I used u=sin^2(9x) and du=cos^2(9x)dx.. When I did the integration, I got u^2/2, which gave me sin^4(9x)/2.. When I evaluated it, it says that I did it wrong..

I, then, realized that I probably should use the Half-angle identities.. But everything just scrambled all over my head and I couldn't understand anything..

Please help me.. Thanks so much!

 

[Deleted member 4993]

It's me again.. >_< I need another help.

Evaluate the definite integral.

sin²(9x)cos²(9x) dx​

​

*from 6 to 13

I tried doing substitution.. I used u=sin^2(9x) and du=cos^2(9x)dx.. incorrect ....When I did the integration, I got u^2/2, which gave me sin^4(9x)/2.. When I evaluated it, it says that I did it wrong..

I, then, realized that I probably should use the Half-angle identities.. But everything just scrambled all over my head and I couldn't understand anything..

Please help me.. Thanks so much!

There are different substitution scheme you can follow. One of those would be:

u = 9x

du = 9 dx

\(\displaystyle \int_6^{13} sin^2(9x)*cos^2(9x) dx \ = \ \int_{54}^{117} sin^2(u)*cos^2(u) \frac{du}{9} \)

\(\displaystyle = \ \frac{1}{36} \int_{54}^{117} sin^2(2u) du \)

\(\displaystyle = \ \frac{1}{72} \int_{54}^{117} [1- cos(4u)] du \)

Now continue....

 

[donnac10]

So do I do one or of them or both of them at the same time?

There are different substitution scheme you can follow. One of those would be:

u = 9x

du = 9 dx

\(\displaystyle \int_6^{13} sin^2(9x)*cos^2(9x) dx \ = \ \int_{54}^{117} sin^2(u)*cos^2(u) \frac{du}{9} \)

\(\displaystyle = \ \frac{1}{36} \int_{54}^{117} sin^2(2u) du \)

\(\displaystyle = \ \frac{1}{72} \int_{54}^{117} [1- cos(4u)] du \)

Now continue....

 

[Deleted member 4993]

So do I do one or of them or both of them at the same time?

I don't understand your question - you have ONE integration to do - what does BOTH mean in this context?

 

[soroban]

Hello, donnac10!

\(\displaystyle \displaystyle \int^{13}_6\sin^2(9x)\cos^2(9x)\,dx\)

We have: .\(\displaystyle \big[\sin(9x)\cos(9x)\big]^2 \:=\:\left[\frac{1}{2}\sin(18x)\right]^2 \)

Then: .\(\displaystyle \displaystyle \tfrac{1}{4}\int^{13}_6\!\!\sin^2(18x)\,dx \;=\; \tfrac{1}{8}\int^{13}_6\big[1 - \cos(36x)\big]\,dx \)

. . \(\displaystyle =\;\frac{1}{8}\left[x - \frac{1}{36}\sin(36x)\right]^{13}_6 \)

Now evaluate it . . .

 

[donnac10]

Sorry.. I got confuse with my question, too.. I just realized how you did it.. Sorry about the confusion and thank you for helping me out!

I don't understand your question - you have ONE integration to do - what does BOTH mean in this context?

 

[donnac10]

That was very clear! Thanks so much!

Hello, donnac10!


We have: .\(\displaystyle \big[\sin(9x)\cos(9x)\big]^2 \:=\:\left[\frac{1}{2}\sin(18x)\right]^2 \)

Then: .\(\displaystyle \displaystyle \tfrac{1}{4}\int^{13}_6\!\!\sin^2(18x)\,dx \;=\; \tfrac{1}{8}\int^{13}_6\big[1 - \cos(36x)\big]\,dx \)

. . \(\displaystyle =\;\frac{1}{8}\left[x - \frac{1}{36}\sin(36x)\right]^{13}_6 \)

Now evaluate it . . .