(After Test 1)
Topics:
• 
This is found by a clever trick that lets us use substitution: see here. (Just the first paragraph.) A similar trick will give the indefinite integral of 
. Once we computed these, we add them to our list of integration formulas.
• More integration using Trigonometric substitution: we carried out some indefinite integrals, but because of the short class period there was no time to finish the last step, which is substituting back for 
in terms of 
. For some indefinite integrals this is rather straightforward, but not for the two we looked at today.
Before going on to that last step, it might be a good idea to brush up on your right-triangle trigonometry definitions of sine, cosine, tangent, secant, cosecant, and cotangent. We’ll use little right triangles and the Pythagorean Theorem to find the formulas for substituting back in the indefinite integrals.
My notes are unavoidably delayed… I hope this is the last delay! But in the meantime here are some goodies:
I am linking three examples (four videos, because one of them got cut off before it was over) from PatrickJMT which show trigonometric substitutions similar to my examples. Please view them and pay special attention to the last part when he draws the right triangles and uses them to find the back-substitutions. For each video I’ve indicated the time marker when that part of the process begins.
Unfortunately the last two examples are only available on YouTube and not on PatrickJMT’s website. Sometimes other videos will start autoplaying on YouTube after the linked one finishes. Sorry, I don’t know how to prevent this from happening.
I recommend you view these in the order I have them below, despite being out of numerical order.
Here is an example, from his website: 
Example 2
The use of the right triangle to find the back-substitution starts at about 5:15 on the video.
Here is another example on YouTube: 
Example 1
The use of the right triangle to find the back-substitution starts at about 14:09 on the video.
Here is another example on YouTube, confusingly also called Example 1 : 
Example 1 part 1, Example 1 part 2
The use of the right triangle to find the back-substitution starts at about 11:50 on part 1 and continues to part 2 (which may autoplay when part 1 is finished).
This video is not as polished as his others, which is probably why it didn’t end up on his website.
Homework:
• Study at least the first two of the videos linked above, paying special attention to the use of the right triangles in substituting back after the integration. Make sure that you understand how the right triangle is constructed and used. (You can also read my notes once they are posted of course!)
• Do the WeBWorK: there is one old and one new assignment. For the problems in this new assignment, Trigonometric substitution 2, you are told which substitution to use, so you do not have to guess. The challenge is to use the proper right triangle to find the back-substitution at the end.
• There will be a quiz on Monday: the topic will be Trigonometric integrals of the type that includes powers of tangent and/or secant, and also a definite integral which uses trigonometric substitution like the example I worked last time (or the first problem in Trigonometric substitution 1 on WeBWorK).
Don’t forget, if you get stuck on a problem, you can post a question on Piazza. Make sure to give your question a good subject line and tell us the problem itself – we need this information in order to answer your question. And please only put one problem per posted question!
)
, so you want to end up with 
in your integrand in order to have 
.
)![\sin(mx)\sin(nx) = \frac{1}{2}\left[\cos(mx-nx) - \cos(mx+nx)\right]](https://s0.wp.com/latex.php?latex=%5Csin%28mx%29%5Csin%28nx%29+%3D+%5Cfrac%7B1%7D%7B2%7D%5Cleft%5B%5Ccos%28mx-nx%29+-+%5Ccos%28mx%2Bnx%29%5Cright%5D&bg=ffffff&fg=000000&s=0 "\sin(mx)\sin(nx) = \frac{1}{2}\left[\cos(mx-nx) - \cos(mx+nx)\right]")
![\sin(mx)\cos(nx) = \frac{1}{2}\left[\sin(mx-nx) - \sin(mx+nx)\right]](https://s0.wp.com/latex.php?latex=%5Csin%28mx%29%5Ccos%28nx%29+%3D+%5Cfrac%7B1%7D%7B2%7D%5Cleft%5B%5Csin%28mx-nx%29+-+%5Csin%28mx%2Bnx%29%5Cright%5D&bg=ffffff&fg=000000&s=0 "\sin(mx)\cos(nx) = \frac{1}{2}\left[\sin(mx-nx) - \sin(mx+nx)\right]")
![\cos(mx)\cos(nx) = \frac{1}{2}\left[\cos(mx-nx) + \cos(mx+nx)\right]](https://s0.wp.com/latex.php?latex=%5Ccos%28mx%29%5Ccos%28nx%29+%3D+%5Cfrac%7B1%7D%7B2%7D%5Cleft%5B%5Ccos%28mx-nx%29+%2B+%5Ccos%28mx%2Bnx%29%5Cright%5D&bg=ffffff&fg=000000&s=0 "\cos(mx)\cos(nx) = \frac{1}{2}\left[\cos(mx-nx) + \cos(mx+nx)\right]")
