## University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson

# Chapter 8 - Section 8.3 - Trigonometric Substitutions - Exercises - Page 439: 7

#### Answer

$\dfrac{25}{2} \sin^{-1}(t/5)+\dfrac{t\sqrt {25 -t^2}}{2} +C$

#### Work Step by Step

Let us consider $t =5 \sin \theta$ and $dt=5 \cos \theta d \theta$ Now, the given integral becomes: $\int \sqrt{25-25 \sin^2 \theta } (5 \cos \theta d \theta ) =\int \sqrt{(25(1-\sin^2 \theta) } (5 \cos \theta d \theta )$ We know that $\cos^2 \theta=1-\sin^2 \theta$ and $\cos^2 \theta =(1/2)+(1/2) \cos 2 \theta$ Then, we have $\int 25 \cos^2 \theta = 25\int (\dfrac{1}{2}+\dfrac{1}{2} \cos ( 2\theta ))d \theta$ Now, integrate and plug in $\sin \theta =t/5 \implies \theta =\sin^{-1}(t/5)$ Thus, $25 (\dfrac{1}{2} \theta+\dfrac{1}{4} \sin ( 2\theta ))d \theta +C=\dfrac{25}{2} \theta+\dfrac{25}{4} (2 \sin \theta \cos \theta) d \theta +C$ or, $=\dfrac{25}{2} \sin^{-1}(t/5)+\dfrac{t\sqrt {25 -t^2}}{2} +C$

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