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: 78

Answer

The formula $$\sec x = \frac{\sin x}{\cos x}+\frac{\cos x}{1+\sin x}$$ is indeed true. It is used to calculate that $$\int \sec x dx =\ln|\sec x+\tan x|+c.$$

Work Step by Step

First, we will derive this rule. It is easier to start from the right side and show that it is equal to the left side $$RS=\frac{\sin x}{\cos x}+\frac{\cos x}{1+\sin x} = \frac{\sin x(1+\sin x)}{\cos x (1+\sin x)}+ \frac{\cos^2 x}{\cos x(1+\sin x)}.$$ This gives $$RS = \frac{\sin x(1+\sin x) + \cos^2 x}{\cos x(1+\sin x)} = \frac{\sin x+\sin^2 x+\cos^2 x}{\cos x(1+\sin x)} = \frac{\sin x+1}{\cos x (1+\sin x)},$$ where we used $\sin^2 x+\cos^2x=1.$ Now dividign both the denominator and the numerator by $1+\sin x$ we get $$RS=\frac{1}{\cos x} = \sec x$$ which is equal to the left side so we derived the formula. Now the integral is $$\int \sec x dx = \int \left(\frac{\sin x}{\cos x}+\frac{\cos x}{1+\sin x}\right) dx =\int \frac{\sin x}{\cos x} dx + \int \frac{\cos x}{1+\sin x} = \int \tan x dx +\int\frac{\cos x}{1+\sin x}dx.$$ The first integral is read directly from the table of integrals $$\int\tan x dx = -\ln|\cos x|+c_1.$$ The second integral can be solved using substitution $t=1+\sin x$ and by differentiating $dt=(1+\sin x)'dx=\cos x dx.$ Putting this into the second integral we have $$\int \frac{\cos x}{1+\sin x}dx = \int\frac{dt}{t} =\ln|t|+c_1 =\ln|1+\sin x|+c_2,$$ where in the end we expressed the substitution $t$ in terms of $x$ again. The initial integral is now $$\int\sec x dx = -\ln |\cos x|+c_1+\ln|1+\sin x| +c_2 = \ln\frac{|1+\sin x|}{|\cos x|}+c,$$ where we simply denoted $c=c_1+c_2$ the new arbitrary constant $c$. Now we will just transform the previous expression to get the formula from the problem: $$\int\sec x dx = \ln\frac{|1+\sin x|}{|\cos x|}+c = \ln\left|\frac{1+\sin x}{\cos x}\right|+c= \ln\left|\frac{1}{\cos x}+\frac{\sin x}{\cos x}\right|+c = \ln|\sec x +\tan x| +c. $$ Where we used that $\frac{1}{\cos x}=\sec x$ and $\frac{\sin x}{\cos x} = \tan x.$
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