Calculus 10th Edition

Published by Brooks Cole
ISBN 10: 1-28505-709-0
ISBN 13: 978-1-28505-709-5

Chapter 8 - Integration Techniques, L'Hopital's Rule, and Improper Integrals - 8.1 Exercises - Page 513: 77

Answer

We find that $a=\sqrt2$ and $b=\frac{\pi}{4}.$ The integral is $$\int\frac{dx}{\sin x+\cos x}=-\frac{\sqrt2}{2}\ln|\csc (x+\pi/4)+\cot (x+\pi/4)|+c.$$

Work Step by Step

We will use the trigonometric identity $\sin x+\cos x=a \sin x\cos b+a\cos x\sin b.$ Coefficients multiplying $\sin x$ and $\cos x$ have to be equal on both sides of the equality: $a\cos b=1$ and $a\sin b=1.$ We will firstly square both of them and add them together: $a^2\cos^2b+a^2\sin^2b=2\Rightarrow a^2(\cos^2b+\sin^2b)=2\Rightarrow a^2=2\Rightarrow a=\sqrt2,$ where we used that $\cos^2b+\sin^2b=1.$ Now dividing both of them we get: $$\frac{a\sin b}{a\cos b}=1\Rightarrow\tan b=1\Rightarrow b=\arctan1=\frac{\pi}{4}.$$ Using the calculated results we get $\sin x+\cos x=\sqrt2\sin(x+\frac{\pi}{4}).$ Putting this into the integral we get: $$\int\frac{dx}{\sin x+\cos x}=\int\frac{dx}{\sqrt2\sin(x+\frac{\pi}{4})}=\frac{1}{\sqrt2}\int\frac{1}{\sin(x+\frac{\pi}{4})}dx=\frac{1}{\sqrt2}\int\csc\left(x+\frac{\pi}{4}\right)dx=-\frac{\sqrt2}{2}\ln|\csc( x+\frac{\pi}{4})+\cot (x+\frac{\pi}{4})|+c.$$
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