If you're seeing this message, it means we're having trouble loading external resources on our website.

If you're behind a web filter, please make sure that the domains *.kastatic.org and *.kasandbox.org are unblocked.

### Course: Integral Calculus (2017 edition)>Unit 6

Lesson 6: Trigonometric substitution

# Trig and u substitution together (part 2)

More of all of the substitution! Created by Sal Khan.

## Want to join the conversation?

• Is there anything wrong with just plugging in cosθ for all the u's and then plugging in arcsin(x/3) for all the θ's?
• I did it with the arcsin and i think its correct. Either way its not going to be a pretty answer and i try to avoid having to do all the trig and geometry stuff, especially if its on a test and im rushed for time.
• so, in general, when given an integral where 2 functions are multiplied or divided one should:
1) Check to see if Trig substitution is applicable, if not
2) Check to see if U Substitution is applicable, if not
3) Integrate by parts

would this be the optimal algorithm for calculating integrals of this format or is there a better way?
• Plan A should be look for obvious u sub. Next, try integration by parts. Then see if it can be turned into u sub by trig identities, then try trig substitution, finally try partial fraction decomposition.
• Where did the -243 go? You copied it as 243 in the following line.
• He distributed the -1 inside the parentheses.
-243(a - b) = 243(b - a)
• How did the 1/3 become 1/9 when it was put in the radical?
• It became that because we're inserting that value into a square root. If you undo the 1/9 from the square root you should be having the original value 1/3. Similar to have an original value of 4, now when you put that in a square root it should be 16 simply because when you take the square root of that you should be able to get your original value 4. I hope that helps someway for people with similar question.
(1 vote)
• We can use the Pythagorean identity, right? Here's what I've done:
cos θ = u, sin θ = x/3
cos^2 (θ) + sin^2 (θ) = 1
u^2 + (x^2)/9 = 1
u^2 = 1 – (x^2)/9
u = sqrt (1 – (x^2)/9)
which is the same as sir Sal's result.
• you can just to u-sub and avoid trig sub. change x^3 to x and x^2 where x^2 = 9-u...and du/dx = -2x. It saved my sanity.
• At , Sal says that u=cos(theta)=sqrt(1-sin^2(theta)). I thought that sqrt(1-sin^2(theta)) = cos^2(theta)
• You're forgetting to take the square root. As you know, sin²Θ + cos²Θ = 1. Then cos²Θ = 1 - sin²Θ. Now take the square root of both sides of the equation to get Sal's result.
• Couldn't you just solve for θ by taking the inverse sine of both sides of (x/3)=sin(θ) to get θ=arcsin(x/3) and then u=cos(arcsin(x/3))?
(1 vote)
• That creates problems because the domain of arcsin(x/3) is just [-3, 3]. We want this to hold in general, not just in that interval.