help! Integration involving Trig Calculus problem

[jessicafanfan]

Please see the attached image for the problem. I can't seem to figure it out...
Any help is appreciated!
Thanks!

 

[MarkFL]

I would look at integration by parts. Suppose we let:

\(\displaystyle u=\tan^5(2x)\implies du= 10\tan^4(2x)\sec^2(2x)\,dx=10\left(\tan^6(2x)+\tan^4(2x)\right)\,dx\)

\(\displaystyle \displaystyle dv=\tan(2x)\sec(2x)\,dx\implies v=\frac{1}{2}\sec(2x)\)

Can you continue?

 

[jessicafanfan]

I did it and I think I can substitute I in one part of my equation

What do I do with the rest?
Thank you so much!

 

[MarkFL]

What do I do with the rest?
Thank you so much!

Let's let:

\(\displaystyle \displaystyle I_2=\int_0^{\frac{\pi}{6}} \tan^6(2x)\sec(2x)\,dx\)

For starters, we must take into account that we are working with a definite integral, so we would in fact have:

\(\displaystyle \displaystyle I_2=\frac{1}{2}\left[\tan^5(2x)\sec(2x)\right]_0^{\frac{\pi}{6}}-5\int_0^{\frac{\pi}{6}} \tan^6(2x)\sec(2x)\,dx-5\int_0^{\frac{\pi}{6}} \tan^4(2x)\sec(2x)\,dx\)

Now, we should observe at this point that the second term on the RHS is a multiple of \(\displaystyle I_2\) and the third term is a multiple of \(\displaystyle I\), so we may write:

\(\displaystyle \displaystyle I_2=\frac{1}{2}\left[\tan^5(2x)\sec(2x)\right]_0^{\frac{\pi}{6}}-5I_2-5I\)

Now, carry out the indicated computation on the first term on the RHS, and solve for \(\displaystyle I_2\)...what do you get?

 

[jessicafanfan]

Thank you so much!!! I got it!!!

Thanks!

Let's let:

\(\displaystyle \displaystyle I_2=\int_0^{\frac{\pi}{6}} \tan^6(2x)\sec(2x)\,dx\)

For starters, we must take into account that we are working with a definite integral, so we would in fact have:

\(\displaystyle \displaystyle I_2=\frac{1}{2}\left[\tan^5(2x)\sec(2x)\right]_0^{\frac{\pi}{6}}-5\int_0^{\frac{\pi}{6}} \tan^6(2x)\sec(2x)\,dx-5\int_0^{\frac{\pi}{6}} \tan^4(2x)\sec(2x)\,dx\)

Now, we should observe at this point that the second term on the RHS is a multiple of \(\displaystyle I_2\) and the third term is a multiple of \(\displaystyle I\), so we may write:

\(\displaystyle \displaystyle I_2=\frac{1}{2}\left[\tan^5(2x)\sec(2x)\right]_0^{\frac{\pi}{6}}-5I_2-5I\)

Now, carry out the indicated computation on the first term on the RHS, and solve for \(\displaystyle I_2\)...what do you get?