7–84. Evaluate the following integrals.
62. ∫ from 0 to ...

Question

7–84. Evaluate the following integrals.

62. ∫ from 0 to π/2 √(1 + cosθ) dθ

Accepted Answer

Hello. In this video, we are going to be calculating the given definite integral. Now, we are given the integral from 0 toi of the square root of 1 minus cosine theta D theta. Now, the first thing we want to do is we want to go ahead and simplify the integrand, but this integrand is quite complicated to simplify on its own. So instead, let's go ahead and try to rewrite the integrand by using a trigonometric identity. Now, one trigonometric identity we can start with, at least, is going to be the half angle identity for sign. That identity is the following. Sine of theta divided by 2 is equal to positive negative square root of 1 minus cosine theta divided by 2. Notice that we have 1 minus cosine theta underneath the square root function, which is similar to what we have in our integrand, but we need to get rid of this denominator of two. So why don't we try to solve for 1 minus cosine theta underneath the square root within this half angle identity. The first thing we can do is we can square both sides of this equation. That will leave us with sine squared of theta divided by 2, and that is equal to 1 minus cosine theta divided by 2. Multiplying both sides of this equation and reordering will leave us with 1 minus cosine of theta is equal to 2 multiplied by sin squared of theta divided by 2. And next, if we were to take the square root of both sides of this equation, we will be left with 1 minus. Cosaint Theater All underneath the square root is equal to the square root of 2 multiplied by sine squared of theta divided by 2. And we can further simplify the right-hand side to be the square root of 2 multiplied by the absolute value of of theta divided by 2. Now, this is going to be the simplest form for a square root of 1 minus cosine theta, but notice that we were integrating on the bounds from 0 to pi. From 0 to pi, we know that sine is going to be a positive quantity, so we can rewrite the sine function without the absolute values. And so that means that the square root of 1 minus cosine theta will be the square root of 2 multiplied by sine of theta divided by 2. So by using this identity, we can rewrite the integral to be the integral from 0 toi of the square root of 2 multiplied by sine of theta divided by 2 d theta, and moving the square root to in front of the integral as a constant will leave us with the square root of 2 multiplied by the integral of 0 toi of sine theta divided by 2 D theta. Now we can solve this type of integral by using a U substitution. If we allow U to equal to theta divided by 2, then DU will equal to 12 D theta. We don't have a 1/2 coefficient in the integran, so if we multiply the DU statement, both sides by 2, we will have 2 DU is equal to D theta, which is going to be a substitution for D theta, and you will substitute for theta divided by 2. Now, since we are applying a use substitution over a definite integral, we also have to switch our bound boundaries of integration. In order to do this, we will plug in our current boundaries into the use substitution. Plugging in 0 for you is going to give us 0 divided by 2, which is equal to 0. And plugging in ie into the use substitution. Will leave us with pi divided by 2. And so that means that by applying this substitution, we can rewrite the definite integral as the square root of 2 multiplied by the integral from 0 to pi divided by 2 of sine of you. D2DU. And we can further rewrite this integral by moving the two as a coefficient in front of the integrand or in front of the integral. So, all we need to do now is we need to take the antiderivative of the sine function. The antiderivative of sine is going to be negative cosine, so we are left with 2 multiplied by the square root of 2, multiplied by negative cosine of U, and this will be evaluated from 0 to pi divided by 2. Multiplying in the 2 square root of 2 will give us -2 square root of 2, multiplied by cosine of u, evaluated from 0 to pi divided by 2. And now we are going to finish this by using the fundamental theorem of calculus part 2. By plugging in our boundaries, we have -2, square root of 2, cosine of pi divided by 2, plus 2 square root of 2 multiplied by cosine of 0. Cosine at pi divided by 2 simplifies to 0, and that means this whole first half is going to simplify to 0. But cosine at zero simplifies to positive 1, and 1 multiplied by 2 square root of 2 is just going to leave us with 2 square root of 2, which is going to be the solution to the definite integral. So I hope this video helps you in understanding how to approach this problem, and we will go ahead and see you all in the next video.

7–84. Evaluate the following integrals.
62. ∫ from 0 to π/2 √(1 + cosθ) dθ

Key Concepts

Definite Integral

Trigonometric Identities

Substitution Method

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