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# Trig and u substitution together (part 1)

All of the substitution! Created by Sal Khan.

## Want to join the conversation?

• Sal explains this in the first video, Intro to trig substitution, but it's not intuitive to me either so it's worth going over again:
If you have a right triangle with hypotenuse of length a and one side of length x, then:
x^2 + y^2 = a^2 <- Pythagorean theorem
where x is one side of the right triangle, y is the other side, and a is the hypotenuse.
So anytime you have an expression in the form a^2 - x^2, you should think of trig substitution. Now here's where the trig comes in:
Drawing our right triangle:

.............../|
............/...|
........./......|
..a. /.........|
..../...........| x
../.............|
/..)._____|
......... y
sin theta = opposite / hypotenuse
sin theta = x/a
solve for x:
a * sin theta = x
• cant i take
u=sin^3Ө
du=3cos^2ӨdӨ
cos^2ӨdӨ=du/3
and 243ʃu(du/3)
is it correct.. cuz i'm gettin a different answer if i do that!!
• ok this is from 8 years ago, but I'll answer anyways lol.

the derivative of your u isn't 3cos^2(x)dx, it's 3sin^2(x)cos(x)
(I'm using x instead of theta)
• At , why can we simply assume that cos(theta) is positive?
• At , when we are taking the integral of the function, shouldn't the result be also multiplied by "u", as the integral of "du" is just "u+C"?

Thank you!
• That is built into the concept of an integral, so no, because you're already doing that.
• Why did he substitute u=cos(theta) instead of u=sin(theta)?
• Sal created an extra sin(theta) so he could use it as du. Sin(theta) is the derivative of -cos(theta). Therefore u must be cos(theta).
• it will work with u-substitution directly with u = sqrt ( 9 - x^2 )
• if u= sqrt (9-x^2), then how does one account for du = 2x without an x in the numerator?
• Sal: You're not getting very far with u-substitution.
Me: You're definitely not trying hard enough! With proper u-substitution, you can solve this problem in a smaller number of steps and do fewer mistakes. Substitute
u^2 = 9 - x^2x du = -u du
The integral thus becomes,
∫u^4 - 9u^2 du
How is this not easier to solve? :D