True or false 1dx9 – x21dtV1 – t2

Given integration to prove or disprove: ∫-3319-x2dx=∫-1111-t2dt

Answered: 1 1 dt 1V1 – t2 9 – x2 xp:

Calculus: Early Transcendentals

8th Edition

ISBN:9781285741550

Author:James Stewart

Publisher:James Stewart

Chapter1: Functions And Models

Section: Chapter Questions

Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...

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True or false

Expert Solution

Step 1

Given integration to prove or disprove:

$\int_{- 3}^{3} \frac{1}{\sqrt{9 - x^{2}}} d x = \int_{- 1}^{1} \frac{1}{\sqrt{1 - t^{2}}} d t$

Step 2

LHS = $\int_{- 3}^{3} \frac{1}{\sqrt{9 - x^{2}}} d x$

LHS = $\int_{- 3}^{3} \frac{1}{\sqrt{3^{2} - x^{2}}} d x$

if expression is $\sqrt{a^{2} - x^{2}}$ then substitute x = a sin t into the expression, therefore, substitute x = 3 sin t and dx = 3 cos t dt into the expression:

$x = 3 \sin t D i f f e r e n t i a t e t h e b o t h s i d e o f t h e e q u a t i o n : \frac{d x}{d t} = 3 \cos t d x = 3 \cos t d t$

$L H S = \int_{- 3}^{3} \frac{3 \cos t d t}{\sqrt{3^{2} - {(3 \sin t)}^{2}}} L H S = \int_{- 3}^{3} \frac{3 \cos t d t}{\sqrt{9 - 9 {(\sin t)}^{2}}} L H S = \int_{- 3}^{3} \frac{3 \cos t d t}{\sqrt{9 (1 - {(\sin t)}^{2}})} L H S = \int_{- 3}^{3} \frac{3 \cos t d t}{3 \sqrt{\cos^{2}} t} {1 - \sin^{2} t = \cos^{2} t} L H S = \int_{- 3}^{3} \frac{c o s t d t}{\cos t} L H S = \int_{- 3}^{3} 1 d t L H S = {[t]}_{- 3}^{3} S i n c e x = 3 \sin t, t h e r e f o r e t = \sin^{- 1} \frac{x}{3} : L H S = {[\sin^{- 1} \frac{x}{3}]}_{- 3}^{3} L H S = \sin^{- 1} \frac{3}{3} - \sin^{- 1} \frac{- 3}{3} L H S = \sin^{- 1} 1 - \sin^{- 1} - 1 L H S = \sin^{- 1} \sin \frac{π}{2} - \sin^{- 1} \sin (- \frac{π}{2}) L H S = \frac{π}{2} - (- \frac{π}{2}) L H S = \frac{π}{2} + \frac{π}{2} L H S = π$

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