Evaluate the following integral int0fracpi6 fraccos 4 xsqrt1...

Name: evaluate the following integral int0fracpi6 fraccos 4 xsqrt1 sin x d x int0fracpi6 fraccos 4 xsqrt1 sin x d x type an exact answer 20956
Uploaded: 2024-09-17T02:35:28-08:00
Duration: 7 min 30 s
Channel: Ma. Theresa Alin
Description: evaluate the following integral int0fracpi6 fraccos 4 xsqrt1 sin x d x int0fracpi6 fraccos 4 xsqrt1 sin x d x type an exact answer 20956

Evaluate the following integral. $$ \int_{0}^{\frac{\pi}{6}} \frac{\cos ^{4} x}{\sqrt{1-\sin x}} d x $$ $$ \int_{0}^{\frac{\pi}{6}} \frac{\cos ^{4} x}{\sqrt{1-\sin x}} d x = $$ (Type an exact answer.)

Transcript

00:01 For this problem, you want to evaluate the integral from 0 to pi over 6 of 2 cosine of x raise of the 4th power all over the square root of 1 minus sine of x with respect to x.

00:11 Now for this, we need to rewrite our function.

00:15 We have to multiply it by the square root of 1 plus sine of x.

00:20 So this becomes the integral from 0 to pi over 6, the 2 cosine of x raised to 4th power times the square root of 1 .000.

00:30 Plus sine of x over the square root of one minus sine of x times the square root of one plus sine of x d x so that'll be interval from 0 to pi over 6 to cosine of x to cosine of x times the square root of 1 plus sine of x all over the square root of 1 minus sine squared of x dx but then in trigonometric identities if we have cosine squared of x plus sine squared of x that equals 1 so cosine squared of x is 1 minus sine squared of x then from here we will have 0 or integral from 0 to pi over 6 to cosign of x 3 to the 4 is power times a square 1 plus sine of x all over the square root of cosine squared of x that's going to equal integral from 0 to pi over 6 to cosine of x to the fourth power times the square root of 1 plus sine of x all over cosine of x d x that will simplify into integral from 0 to pi over 6 to cosine cubed of x times the square root of 1 plus sine of x d x as the same as the integral from 0 to pi over 6 to cosine squared of x times the square root of 1 plus sine of x times cosine of x d x but cosine squared of x if we recall the identity from before cosine squared of x is 1 minus sine squared of x and that becomes the integral from 0 to pi over 6 2 times 1 minus sine squared of x times the square of x times the square root of 1 plus sine of x times cosine of x d x and then we're going to do substitution and here we want to set u equal to sign of x so d u equals cosine of x d x and if x is zero u equals u equals 0 equals 0 if x is pi over 6 we have u equal to one half.

02:55 And then from here we will have the integral from 0 to 1 half of 2 times 1 minus u squared times the square root of 1 plus u d u.

03:07 And then we know 1 minus u squared is the same as 1 minus u times 1 plus u.

03:18 So if we multiply this to the square root of 1 plus u, 1 minus u squared times square root of 1 plus u is the same as 1 minus u times 1 plus u times the square root of 1 plus u, we can combine these.

03:36 That's 1 minus u times 1 plus u raised to 3 over 2.

03:42 And this will become two integral from 0 to 1 half of 1 minus u times 1 plus u raised to 3 over 2, du.

03:53 And then we're going to do substitution again.

03:55 This time we want to set v equal to 1 plus u.

03:58 1 plus u...