Trigonometric substitutions Evaluate the following integrals using trigonometric substitution. ∫

Trigonometric substitutions Evaluate the following integrals...

Key Concepts

Change of Variable Technique

The change of variable technique is inherent in trigonometric substitution, where the original variable is replaced with a new variable related to an angle. This not only transforms the integral into a potentially simpler one but also requires adjusting the differential accordingly. Once the integral is computed with respect to the new variable, the result is converted back into the original variable, completing the process.

Trigonometric Substitution

Trigonometric substitution is a method used to simplify integrals containing expressions like ?(a² + b²x²), ?(a² - b²x²), or ?(b²x² - a²) by substituting x with a trigonometric function (such as x = (1/b)tan? or x = (1/b)sin?). This substitution leverages trigonometric identities to transform complex algebraic expressions into trigonometric forms that are easier to integrate.

Use of Pythagorean Identities

Pythagorean identities, for example, 1 + tan²? = sec²? and 1 - sin²? = cos²?, are essential in trigonometric substitution. They allow the replacement of quadratic expressions under a square root with trigonometric functions, thereby simplifying the integrand into a form that can be readily integrated.

Transcript

00:01 Section 8 .4, number 41, we're asked to perform the integral 1 over 1 plus 4 x squared to the 3 halves using trig substitution.

00:09 So most of our trig substitution, all of them, there of the form x squared plus a squared, x squared minus a squared, a squared minus x squared.

00:16 So i've got to work with this integral before i make my substitution.

00:21 So if i look at, let's just look at this 1 plus 4x squared to the 3 halves.

00:29 Well i could rewrite this let's just factor a four out so four times one -fourth plus x squared to the three -haves okay and then that is just what four to the three -halfs times one -fourth plus x squared to the three - halves and so this is just simply four to the three -haves is eight so this is eight times one -fourth plus x squared to the three halves.

01:06 So now i'm in the form where i can use my trick substitution with a tangent substitution.

01:12 So we can rewrite this integral.

01:14 This is the integral of one -eighth, one over, so it's one -fourth plus x squared to the three - halves dx.

01:30 Now my substitution here is going to be x equal one -half tangent of theta.

01:39 Then dx is going to be one -half secant squared theta d -theta and then one -fourth plus x squared to the three -halfs is going to be one -fourth plus when you square x you're going to get one -fourth tangent squared all of this to the three -halfs so that's one -fourth one -plus tangent squared which is sequent squared so this is one -fourth siquant squared of theta to the three halves and this is just simply one half siquant cubed of theta did i do that right so sorry no let me back up it was good into the last step here so try not to make that mistake one half to the three halves power so that's gonna be one half cube this is one eighth so one eighth siquant cubed of theta so this integral transforms into one eighth and then that term one -fourth plus x squared to the three halves we just determine what that is that is one -eighth so that's going to be one -eighth of the siquant cubed of theta and then d x dx is one -half secant squared of theta d theta and so what you see here is in terms of cancellation you've got one eighth divided by one eighth and you've got secant squared over secant cubed and so this just simply turned into this is one -half the integral one over the secant of theta d -theta so you can rewrite that is one -half integral of the cosine of theta d -theta.

04:12 So now we're in a much easier place for integration.

04:15 So this answer is just going to be what one -half, integrate the cosine function to get the sign, so sine of theta, plus a constant of integration...

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