Evaluate the integrals by making a substitution (possibly trigonometric) and then applying a red

step 1 Okay, so I'm going to be solving this definite interval right here. So to start with, we're goin

Evaluate the integrals by making a substitution (possibly...

Key Concepts

Substitution

Substitution is an integration technique where the variable of integration is replaced to simplify the integral. It involves defining a new variable that transforms the integrand into a more manageable form, making it easier to evaluate the integral.

Trigonometric Substitution

Trigonometric substitution is a method used to simplify integrals that involve expressions like square roots or quadratic forms. By substituting a trigonometric function for the original variable, the integral can often be transformed into one that leverages trigonometric identities, thereby simplifying the computation.

Reduction Formula

A reduction formula is a recursive relationship that expresses an integral with a power or a complex function in terms of a similar integral with a lower power or simpler function. This technique reduces the complexity of the original integral by repeatedly lowering the exponent or index until the integral can be directly evaluated.

Definite Integral Evaluation

Definite integral evaluation involves calculating the net area under a curve between specified limits. When using substitution methods, the integration limits must also be transformed accordingly, ensuring the integral is correctly evaluated over the new variable’s interval.

Transcript

00:02 Okay, so i'm going to be solving this definite interval right here.

00:06 So to start with, we're going to solve the integral first and then apply the boundaries after.

00:13 So the first step that we're going to do is use a trigonometric substitution.

00:19 We're going to let y be equal to sine theta.

00:30 This means that if we differentiate it, we get ey is equal to cosine.

00:36 Theta d theta.

00:41 Also, we know that 1 minus y squared is equal to 1 minus sine squared theta.

00:57 Then from our trigonometric identities, we know that this is equal to cos squared theta.

01:10 So this means that we've got the integral.

01:15 Now i'm going to so does not apply these boundaries for now.

01:20 We're not even about just for now because they're annoying to write down all the time.

01:26 But we've got cosine theta b theta over cosine squared theta to the power of five paths.

01:49 So this bottom part here that we do with the exponent, we multiply it through with this exponent.

01:57 And that will make it cosine to the power of 5.

02:01 Then we also have a cosine in the numerator, which means we can cancel that and cancel these exponents.

02:10 What we are left with is a cosine to the power of 4 of theta.

02:16 So now our integral looks like this, d theta over cosine to the power of 4 of theta.

02:34 Now what we can do is we can apply a reduction formula.

02:39 The reduction formula that we're going to apply is the integral of b theta over cosine to the power of some exponent n of a times theta is equal to sign of a theta divided by a times cosine to the power of a times cosine to the power of n minus 1 of a theta times 1 over n minus 1 plus n minus 2 over n minus 1 times the integral of d theta over cosine to the power of n minus 2 of a theta so in this case you have n equals 4 and a equals 1.

04:04 So applying that, applying those values to this formula, what we get is sine theta over cosine cubed theta times 1 over 4 minus 1 plus 4 minus 2 over 4 minus 2 over 4 minus 1 times the integral of e theta over cosine squared theta.

04:53 So we now have a simpler integral to deal with the cosine to the power of 2 instead of our original cosine to the power of 4.

05:02 So what we can do now is just simplify this side a little bit.

05:08 This is a one -third factor.

05:10 And if we can hold the cosine into the numerator by changing it to a two.

05:16 Secant theta, secant cubed theta instead of cosine in the numerator and this factor right here becomes 2 over 3 so we now have 1 third times sine theta times secant cubed theta plus 2 thirds times the integral now, what we're going to do here is actually change this into secad square theta.

06:03 So this is ccad squared theta d theta because ccd theta is equal to 1 over cosine theta.

06:17 So that makes this integral the integral of ccad squared theta, d theta.

06:30 So we know that from our common derivatives, the derivative of tan theta is equal to secan squared theta.

06:49 So if we integrate this, which is what we're doing here, what we get is tan theta.

06:58 We know this just from our common derivatives.

07:02 So we can then write this whole solution in terms of the you now have one third sign theta times secant cubed theta plus two thirds times tan theta so what we can do now is we can apply our original boundaries to our solution in terms of theta by adjusting these in terms of balance of theta or we can reconvert back into y and then apply these regular boundaries as they are what i'm going to do is convert back into terms of y and then apply the original boundaries but either way it would be correct so we know that y is equal to sine theta and sine data is equal to opposite over adjacent sorry opposite over hypotenuse so if we have a right triangle here with this angle being theta, we can say that this side is y and this side is 1.

08:27 So sine theta is opposite or adjacent, or opposite or hypotenuse, sorry, which is y over 1, which is equal to y.

08:36 Then applying epithgowian theorem, you can say that this side is equal to the square root of 1 minus y square.

08:43 So what we want to know now is sine theta, secant cubed theta, or secantheta, and tan theta.

08:56 So secant data is equal to 1 over cosine theta, and cosine theta is equal to adjacent over hypotenuse, which is equal to the square root of 1 minus y squared over 1.

09:23 That secant theta is equal to 1 over the square root of 1 minus y square.

09:33 Also, 10 theta is equal to opposite over adjacent.

09:41 So this is equal to y over the square root of 1 minus y squared...

Please give Ace some feedback