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Chapter 6: Problem 16

Evaluate the following integrals. Include absolute values only when needed. $$\int_{0}^{\pi / 2} \frac{\sin x}{1+\cos x} d x$$

Short Answer

Expert verified
Question: Calculate the value of the definite integral $\int_{0}^{\pi / 2} \frac{\sin x}{1+\cos x} d x$. Answer: The value of the definite integral is $\frac{\pi}{4}$.

Step by step solution

01

Identify a substitution to simplify the integrand

We can use the following trigonometric substitution: $$t = \tan \frac{x}{2} \Rightarrow \frac{1 - \cos x}{1 + \cos x} = -\frac{1 - t^2}{1 + t^2}$$ This substitution will simplify the integrand and make it easier to find the antiderivative.
02

Compute derivatives of the substitution

Compute \(\frac{d t}{d x}\) to express \(d x\) in terms of \(d t\): $$\frac{d t}{d x} = \frac{1}{2} \cdot \frac{d(\tan x)}{d x} \Rightarrow d t = \frac{1}{2}\sec^2\frac{x}{2} d x$$
03

Substitute and simplify the integrand

Substitute \(t\) and \(d x\) in the integral and simplify it: $$\int_{0}^{\pi / 2} \frac{\sin x}{1+\cos x} d x = \int_{0}^{1} \frac{2t}{1-t^2} \frac{2}{1 + t^2} d t$$
04

Find the antiderivative

Now, find the antiderivative of the simplified integrand: $$\int \frac{4t}{(1 - t^2)(1 + t^2)} d t$$ Perform partial fraction decomposition, and then integrate each term: $$\int \frac{4t}{(1 - t^2)(1 + t^2)} d t = \int \left(\frac{A}{1 - t^2} + \frac{B}{1 + t^2}\right) d t$$ By solving for coefficients A and B, we find that \(A = -1\) and \(B = 1\), so the integral is now: $$\int \left(-\frac{1}{1-t^2} + \frac{1}{1+t^2}\right) dt = -\int\frac{1}{1-t^2}dt + \int\frac{1}{1+t^2}dt$$ Now we can find the antiderivative: $$-\int\frac{1}{1-t^2}dt + \int\frac{1}{1+t^2}dt= -\frac{1}{2}\ln|1 - t^2| +\tan^{-1}(t) + C$$
05

Evaluate the integral for given limits

Evaluate the integral for limits 0 and 1: $$\int_{0}^{\pi / 2} \frac{\sin x}{1+\cos x} d x = -\frac{1}{2}\ln|1 - t^2| +\tan^{-1}(t) \Bigg|_0^1 = -\frac{1}{2}\ln|1 - 1| + \tan^{-1}(1) - 0 = \frac{\pi}{4}$$ Thus, the final answer is: $$\int_{0}^{\pi / 2} \frac{\sin x}{1+\cos x} d x = \boxed{\frac{\pi}{4}}$$

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Most popular questions from this chapter

Use l'H么pital's Rule to evaluate the following limits. \(\lim _{x \rightarrow 1^{-}} \frac{\tanh ^{-1} x}{\tan (\pi x / 2)}\)

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