Calculus 8th Edition

Published by Cengage
ISBN 10: 1285740629
ISBN 13: 978-1-28574-062-1

Chapter 4 - Integrals - 4.5 The Substitution Rule - 4.5 Exercises - Page 346: 39


$$\int_0^{\pi/6}\frac{\sin t}{\cos^2t}dt=\frac{2}{\sqrt3}-1$$

Work Step by Step

To evaluate the integral $$\int_0^{\pi/6}\frac{\sin t}{\cos^2t}dt$$ we will use substitution $\cos t=z$ which gives us $-\sin tdt=dx\Rightarrow \sin tdt=-dz$ and the integration bounds would be: for $t=0$ we have $z=1$ and for $t=\pi/6$ we have $z=\sqrt 3/2$ so we get: $$\int_0^{\pi/6}\frac{\sin t}{\cos^2t}dt=\int_1^{\sqrt3/2}-\frac{dz}{z^2}=-\left.\frac{z^{-1}}{-1}\right|_{-1}^{\sqrt3/2}=\left.\frac{1}{z}\right|_{-1}^{\sqrt3/2}=\frac{1}{\sqrt3/2}-\frac{1}{1}=\frac{2}{\sqrt3}-1$$
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