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Chapter 5: Problem 42

Use a change of variables to evaluate the following definite integrals. $$\int_{0}^{\pi / 4} \frac{\sin x}{\cos ^{2} x} d x$$

Short Answer

Expert verified
Question: Evaluate the following definite integral: $$\int_{0}^{\pi / 4} \frac{\sin x}{\cos ^{2} x} d x$$ Answer: The definite integral evaluates to $$-\sqrt{2} + 1$$.

Step by step solution

01

Choose a substitution

Choose \(u = \cos x\). This substitution is useful because the denominator of the integrand will simplify into a power of \(u\), while the numerator will have a term that can be easily replaced using the differential of \(u\).
02

Find the differential

Differentiate \(u\) with respect to \(x\): $$\frac{d u}{d x} = -\sin x$$ Now, find the differential \(dx\) in terms of \(du\): $$d x = -\frac{d u}{\sin x}$$
03

Rewrite the integral

Substitute the chosen function and its differential into the integral: $$-\int_{0}^{\pi / 4} \frac{\sin x}{u^{2}}(-\frac{d u}{\sin x})$$ Simplify the integral, $$\int_{0}^{\pi / 4} \frac{1}{u^{2}} d u$$
04

Change the limits of integration

Next, we need to change the limits of integration to match the new variable \(u\). Use the substitution \(u=\cos x\) to find the new limits of integration: Lower limit: \(u(\frac{0}{\pi}) = \cos 0 = 1\) Upper limit: \(u(\frac{\pi}{4}) = \cos \frac{\pi}{4} = \frac{1}{\sqrt{2}}\) So, the new integral becomes, $$\int_{1}^{\frac{1}{\sqrt{2}}} \frac{1}{u^{2}} d u$$
05

Evaluate the integral

Now, we can evaluate the integral: $$\int_{1}^{\frac{1}{\sqrt{2}}} \frac{1}{u^{2}} d u = \left[-\frac{1}{u}\right]_{1}^{\frac{1}{\sqrt{2}}}$$ Evaluate the antiderivative at the limits of integration: $$-\frac{1}{\frac{1}{\sqrt{2}}} + \frac{1}{1} = -\sqrt{2} + 1$$ So, the definite integral is $$\int_{0}^{\pi / 4} \frac{\sin x}{\cos ^{2} x} d x = -\sqrt{2} + 1$$

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Most popular questions from this chapter

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