91Ó°ÊÓ

Derive the formula for \(\int \sec x \, d x\) in the following two ways: (a) By writing $$\begin{aligned} \frac{1}{\cos x} &=\frac{\cos x}{\cos ^{2} x} \\ &=\frac{\cos x}{1-\sin ^{2} x} \\ &=\frac{1}{2}\left[\frac{\cos x}{1+\sin x}+\frac{\cos x}{1-\sin x}\right], \end{aligned}$$ an expression obviously inspired by partial fraction decompositions. Be sure to note that \(\int \cos x /(1-\sin x) d x=-\log (1-\sin x) ;\) the minus sign is very important. And remember that \(\frac{1}{2} \log \alpha=\log \sqrt{\alpha} .\) From there on, keep doing algebra, and trust to luck. (b) By using the substitution \(t=\tan x / 2 .\) One again, quite a bit of manipulation is required to put the answer in the desired form; the expression \(\tan x / 2\) can be attacked by using Problem \(15-9,\) or both answers can be expressed in terms of \(t .\) There is another expression for \(\int \sec x \, d x\) which is less cumbersome than \(\log (\sec x+\tan x) ;\) using Problem 15-9 we obtain $$\int \sec x \, d x=\log \left(\frac{1+\tan \frac{x}{2}}{1-\tan \frac{x}{2}}\right)=\log \left(\tan \left(\frac{x}{2}+\frac{\pi}{4}\right)\right).$$ This last expression was actually the one first discovered. and was due, not to any mathematician's cleverness, but to a curious historical accident: In 1599 Wright computed nautical tables that amounted to definite integrals of sec. When the first tables for the logarithms of tangents were produced, the correspondence between the two tables was immediately noticed (but remained unexplained until the invention of calculus).

Step by step solution

01

Rewrite the integrand

Start by rewriting \( \frac{1}{\cos x} \) using trigonometric identities. \[ \frac{1}{\cos x} = \frac{\cos x}{\cos^2 x} = \frac{\cos x}{1 - \sin^2 x} \]

02

Partial fraction decomposition

Express \( \frac{\cos x}{1 - \sin^2 x} \) as partial fractions. \[ \frac{\cos x}{1 - \sin^2 x} = \frac{1}{2} \left[ \frac{\cos x}{1 + \sin x} + \frac{\cos x}{1 - \sin x} \right] \]

03

Integrate each term

Integrate both terms separately. Use the given information that \( \int \frac{\cos x}{1 - \sin x} dx = -\log (1 - \sin x) \) and note the minus sign: \[ \int \sec x \, dx = \frac{1}{2} \left[ \log (1 + \sin x) - \log (1 - \sin x) \right] \]

04

Simplify the logarithms

Combine and simplify the logarithms: \[ \frac{1}{2} \left[ \log (1 + \sin x) - \log (1 - \sin x) \right] = \frac{1}{2} \log \left( \frac{1 + \sin x}{1 - \sin x} \right) = \log \sqrt{ \frac{1 + \sin x}{1 - \sin x} } \]

05

Substitution method

Use the substitution \( t = \tan \frac{x}{2} \). Recall \( \sec x = \frac{1 + \tan^2 \frac{x}{2}}{1 - \tan^2 \frac{x}{2}} \) and \( dx = \frac{2}{1 + t^2} \). Substituting these in, we simplify.

06

Integrate using substitution

Integrate using the substitution \( t \). The integral transforms to: \[ \int \sec x \, dx = \int \frac{2}{1 - t^2} dt \]

07

Conclusion

Continue integrating and simplifying, yielding: \[ \int \sec x \, dx = \log \left( \frac{1 + \tan \frac{x}{2}}{1 - \tan \frac{x}{2}} \right) = \log \left( \tan \left( \frac{x}{2} + \frac{\pi}{4} \right) \right) \]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Trigonometric integrals

Trigonometric integrals involve the integration of functions that contain trigonometric expressions. When dealing with the integral of the secant function, \( \int \sec x \, dx \), it requires both trigonometric identities and a solid understanding of integral calculus. By breaking down and manipulating trigonometric functions, we can simplify the integration process. In the case presented, rewriting secant as \( \frac{1}{\cos x} \) and using other trigonometric identities like \( \cos^2 x + \sin^2 x = 1 \) helps in decomposing the integrand into simpler parts.

• Recognize trigonometric identities are key to transforming the expression.
• Partial fractions are used in breaking complex fractions into simpler fractions.
• Work term by term to solve the integrals separately before combining them.

This integration approach demonstrates the clever manipulation of integrands using trigonometric identities and partial fractions before directly integrating.

Partial fraction decomposition

Partial fraction decomposition is a technique used to express a complex rational function as a sum of simpler fractions. This is particularly useful in integration where the integrand is a complex fraction. In the example provided, we start with
\[ \frac{\cos x}{1 - \sin^2 x} = \frac{1}{2} \left[ \frac{\cos x}{1 + \sin x} + \frac{\cos x}{1 - \sin x} \right] \]

• First, identify the denominator as a factorable polynomial like \( a^2 - b^2 = (a+b)(a-b) \).
• Express the fraction as a combination of simpler fractions with these new factors in the denominators.
• The aim is to simplify the integrands, which makes it straightforward to integrate them individually.

By applying this method, we reduce the complexity of the integrand, making the integral more manageable to solve.

Substitution method

The substitution method, also known as u-substitution, is a technique used to simplify the integration process. For the integral of the secant function, we use the substitution
\( t = \tan \frac{x}{2} \). This substitution transforms complex trigonometric integrals into algebraic ones. By expressing \( \sec x = \frac{1 + \tan^2 \frac{x}{2}}{1 - \tan^2 \frac{x}{2}} \) and finding the differential
\( dx = \frac{2}{1 + t^2} \), we substitute these into the integral:
\[ \int \sec x \, dx = \int \frac{2}{1 - t^2} dt \]

• First, substitute the trigonometric function with a simpler algebraic variable.
• Transform the integral into an easier form to integrate.
• Integrate and then substitute back the original variable for the final solution.

This greatly simplifies the process, especially for trigonometric integrals that are otherwise challenging.

Logarithmic integration

Logarithmic integration involves integrals that result in logarithmic functions after integration. In the derivation of the integral of the secant function, partial fraction decomposition leads to logarithmic functions. Specifically, we note that:
\[ \int \frac{\cos x}{1 - \sin x} dx = -\log (1 - \sin x) \]
By integrating terms separately and using the relationship between logarithms, we combine and simplify:
\[ \frac{1}{2} \left[ \log (1 + \sin x) - \log (1 - \sin x) \right] = \frac{1}{2} \log \left( \frac{1 + \sin x}{1 - \sin x} \right) = \log \sqrt{\frac{1 + \sin x}{1 - \sin x}} \]

• Recognize integral results that fall into logarithmic forms.
• Apply logarithmic identities to combine and simplify results.
• Understand that logarithmic functions can emerge naturally from integration of fraction forms.

This demonstrates how complex integrals can resolve into simpler logarithmic expressions, helping in understanding their evaluation.