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# More trig sub practice

Example of using trig substitution to solve an indefinite integral. Created by Sal Khan.

## Want to join the conversation?

• Why is the integral of d(theta) = theta?
• All d(theta) means is that you are integrating with respect to theta. It's just like how dx means to integrate with respect to 'x'. An indefinite integral of (1/sqrt[2])dx would be (x/sqrt[2]). So if we took the integral of (1/sqrt[2]) with respect to theta, the answer is just (1/sqrt[2])*theta. Don't forget that (1/sqrt[2]) is a constant. To take the antiderivative of any constant, you just tack on the variable that you are dealing with.
• If I ever got a question like this on my exam (which won't be 2 years from now), how would I go about doing it?
How will it strike me to take 3 common, to convert the question into trig identities, etc etc, so many more things?
Would anyone be able to do a question like this if they saw it for the first time in their lives?
(assuming they are not mathletes like Sal of course)
• The biggest clue that you need to use trig substitution that Sal didn't mention is that the term 1/sqr(3-2x^2) looks very similar to the derivative of arcsin(x), which you've learned as 1/sqr(1-x^2). So naturally, it's safe to guess that the antiderivative of 1/sqr(3-2x^2) is an inverse sine function.

Of course, by the time the average calc student figures that out they will probably be slamming their head on their desk trying to do this problem with u-substitution.
• Someone help me out with the final answer! Arcsine has domain restrictions, how is this accounted for? do you need to simplify it further for a more complete answer? does this mean the value which arcsine is being evaluated at needs to fit within the domain, therefore x needs to be within the domain?
• If I substitute sin^2theta = cos^2theta I get a negative sign in the final result. How can two answers be possible or am I making some mistake?
• it's because arcsinθ = -arccosθ + π/2
See how important the + C is? :)
• At , Sal takes (2/3)x^2 = sin^2(theta).
But shouldn't it be (2/3)x^2 = m.sin^2(theta) where m is any arbitrary constant?
The range of the sine function is only from -1 to 1.
In case our x is greater than, say, 3, take 6 for example, then (2/3)x^2 cannot equal sin^2(theta), since sin^2(theta) can never equal 24, since it can equal 1 at max.
This where the constant 'm' will come in handy.
Why is it okay just to use sin(theta) ?
• You are wrong that the sine maps only to the range [-1,1]. While its true if you are dealing with the real numbers, it is not true if you are dealing with complex numbers. For example, sin(pi/2-i ln(2+sqrt(3)))=2
• does dx mean a very small change in x?
• At , when he takes the square root of the equation, should he not put in a plus or minus (+-), since squaring a positive or a negative will give you the same result?
• It would've gave the same answer in the end.
• When you make the substitution x=a*sin(theta), you are limiting the domain of x from all real numbers to the interval [-a, a]. Wouldn't this mean the answer is only valid for x in the interval [-a, a]? Also, how can we assume that sqrt(sin^2 x) = sin x and not -sin x?
• EXCELLENT QUESTION!
I am pretty sure Sal deals with these nuances of trig sub in other videos.

We can make the trig substitution x = a sin θ provided that it defines a one-to-one function. This can be accomplished by restricting θ to lie in the interval [-π/2, π/2] (for cos and sin).

The point of trig sub is to get rid of a square root, which by its very nature also has a domain restriction. If we change the variable from x to θ by the substitution x = a sin θ, then we can use the the trig identity 1 - sin²θ = cos²θ which allows us to get rid of the square root sign, since:

√(a² - x²) = √(a² - a²sin²θ) = √(a²(1 - sin²θ)) = √(a²cos²θ) = a|cosθ|

And since θ is in [-π/2, π/2], we can say that a|cosθ| = acosθ.
• why is 1/sqrt2*S d theta = 1/sqrt2 theta + C?????
• S d(theta) = is the same thing as = S 1 d(theta) and the antiderivative of one is theta. because your taking the antiderivative with RESPECT to theta.
• how do you solve for theta like sal did in ?