3.3 Trigonometric Substitution

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Instructor: The last one we have in this-- dx over the square root of x squared minus 4. And notice, that x is greater than 2. Our a value is 2. So we are still going to use the positive version of that square root when we get there. Now, we will make our substitution of x equals 2 secant theta dx equals 2, secant theta tangent theta d theta. So our substitution here is for dx 2 secant theta tangent theta divided by d theta, divided by the square root of-- squaring that is 4 secant squared minus 4, which will become a tangent.

So that will be the integral of 2 secant theta tangent theta d theta, over 2 tangent theta. That will be just a secant. Integral of secant theta d theta, which is going to be a natural log absolute value of secant theta plus tangent-- we need an absolute value there-- plus c. Now, we need to determine-- based on a triangle-- what these are going to become. So secant is 1 over cosine. So this is hypotenuse over adjacent.

So hypotenuse over adjacent-- so this the square root of 4 minus x squared. So secant-- well, secant is x over 2. So the absolute value of that. Plus-- now, tangent is opposite over adjacent. So this is the square root 4 minus x squared, square root over 2. So value plus c.

[ENDS AT 38:32]