Problem 94 Hard Difficulty

(a) If $ f $ is continuous, prove that
$$ \int^{\pi/2}_0 f(\cos x) \,dx = \int^{\pi/2}_0 f(\sin x) \,dx $$

(b) Use part (a) to evaluate $ \displaystyle \int^{\pi/2}_0 \cos^2 x \,dx $ and $ \displaystyle \int^{\pi/2}_0 \sin^2 x \,dx $

Answer

a. $=\int_{0}^{\pi / 2} f(\sin x) d x$
b. $\int_{0}^{\pi / 2} \cos ^{2} x d x=\int_{0}^{\pi / 2} \sin ^{2} x d x=\frac{\pi}{4}$

Related Courses

Calculus 1 / AB

Calculus: Early Transcendentals

Chapter 5

Integrals

Section 5

The Substitution Rule

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Video Transcript

proof is to ICO and use that we value is the problem. So here we first observed that there's a should not Dante Cose I x is a sem aside pie over to minus X. That means thiss should be f off side pi minus two over minus X. Here we use u substitution. You know, most pie over to minus Next you is most active. Thanks. Way open another page and see what it gives us. So U equals pi over to minus X So excess, Cyril, you is you Is pi over too? When excess pirate too use you and this F off sign of pie over to minus X becomes f off. Sign you and the ex It is the CMAs negative to you. Sorry. Here, here. We know if we swapped after bond or bomb. Which way Beauty Private. Another negative sign. So this because to this a U N ex actress name of variables like right this So we start with we'LL start with the formula on the left hand side and and we got the formula on the right hand side. So we prove the identity which is the first part They use the first part to evaluate this integral oil On the other page here it looks like if we pick x square f of X equals x square, then we'LL have It's a girl from zero to pie over too Sorry, we used part and pig If off because x squared then in this case it keeps us by the conclusion ofthe party We know that this and this it's a simple you look at the formula here f f it's x squared that eleven size coz I square the right hand side sine squared So sorry, here's not pie should be pi over too And that said this too high Is there a easy way to compute the eye? Well, we know coz I square is is one minus sai square So this also echoes zero to pi over two one minus zero two pi over too size square Next e X which far definitional Hi! And this are supported the Grover Constant because this is a pie over two minus I So here we got the equation. You know, we get the equation. Hi, because time over, it's you minus side, which means I course two eyeholes pyre or two. So I host Pi over four and that is our final interviews with Defi i Toby Deceit Group

Frank L.

Related Courses

Calculus 1 / AB

Calculus: Early Transcendentals

Chapter 5

Integrals

Section 5

The Substitution Rule

Problem 94 Hard Difficulty

(a) If $ f $ is continuous, prove that
$$ \int^{\pi/2}_0 f(\cos x) \,dx = \int^{\pi/2}_0 f(\sin x) \,dx $$

(b) Use part (a) to evaluate $ \displaystyle \int^{\pi/2}_0 \cos^2 x \,dx $ and $ \displaystyle \int^{\pi/2}_0 \sin^2 x \,dx $

Answer

a. $=\int_{0}^{\pi / 2} f(\sin x) d x$
b. $\int_{0}^{\pi / 2} \cos ^{2} x d x=\int_{0}^{\pi / 2} \sin ^{2} x d x=\frac{\pi}{4}$

Related Courses

Calculus: Early Transcendentals

Related Topics

Discussion

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Calculus 1 / AB Courses

Precalculus Review - Intro

Functions on the Real Line - Overview

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Watch More Solved Questions in Chapter 5

Video Transcript

Calculus: Early Transcendentals

Catherine R.

Heather Z.

Kristen K.

Michael J.

Precalculus Review - Intro

Functions on the Real Line - Overview

a. $=\int_{0}^{\pi / 2} f(\sin x) d x$b. $\int_{0}^{\pi / 2} \cos ^{2} x d x=\int_{0}^{\pi / 2} \sin ^{2} x d x=\frac{\pi}{4}$

Calculus: Early Transcendentals

Catherine R.

Heather Z.

Kristen K.

Michael J.

Precalculus Review - Intro

Functions on the Real Line - Overview

Catherine R.

Heather Z.

Kristen K.

Michael J.

a. $=\int_{0}^{\pi / 2} f(\sin x) d x$
b. $\int_{0}^{\pi / 2} \cos ^{2} x d x=\int_{0}^{\pi / 2} \sin ^{2} x d x=\frac{\pi}{4}$