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## Integral Calculus (2017 edition)

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

Lesson 6: Trigonometric substitution

# 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)dx would be (x/sqrt). So if we took the integral of (1/sqrt) with respect to theta, the answer is just (1/sqrt)*theta. Don't forget that (1/sqrt) is a constant. To take the antiderivative of any constant, you just tack on the variable that you are dealing with.
• 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? • 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? • 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.
• how do you solve for theta like sal did in ? • how the heck do you factor out 3 from 3-2x^2? first time i heard of this...... • It is a great algebraic tool and a very useful one.
If you have 3 - 2x² that can be rewritten as 3(1- ⅔x), since when you multiply it out you get:
3(1) - (3)(⅔)x. You can see how the 3 cancels with the 3 in the denominator of ⅔, so that you get the original expression back - thus the expressions are equivalent.

You will see insights and algebraic manipulations like this more and more, espcially in integral calculus as we often need to transform an expression into another equivalent expression that we can find the integral for, just like was done in the video.
• I'm having trouble understanding the trigonometric substitutions.
By substituting a x=sin(a) , aren't we limiting the values that x can take? Like sin(a) can only be a value between -1 to 1. So we are limiting the values of x to also be between -1 and 1. Isn't that saying we can not take any value of x other than those between these limits? And if so how is the substitution useful in the general case of solving a problem? • That's a very good question, and lots of people wonder the same thing.

If you look at the problem Sal is solving in this video, you'll see that the values of x already are limited. In particular, x has to be between ±√(3/2). If it's beyond that, then the part under the radical in the original equation will be negative.

When Sal finishes, you'll notice that the part he ends up with has the exact same limitation: x still only can be between ±√(3/2). If it's beyond that, then you'd leave the domain of arcsin.

Take a look at the other videos in this section (on trig substitution), and you'll notice that they all have similar limits from the outset. 