Question

$$\text { Evaluate } \int \frac{d x}{1+\sin x+\cos x} \text { using the }$$ substitution \(x=2 \tan ^{-1} \theta .\) The identities \(\sin x=2 \sin \frac{x}{2} \cos \frac{x}{2}\) and \(\cos x=\cos ^{2} \frac{x}{2}-\sin ^{2} \frac{x}{2}\) are helpful.

Short answer

Answer: The integral is $$\ln |1+(\tan\frac{x}{2})^2 +2\tan\frac{x}{2}|+C$$.

Step-by-step solution

Find \(\frac{dx}{d \theta}\)

Let \(x=2 \tan^{-1}\theta\). Now, we need the derivative of \(x\) with respect to \(\theta\). Differentiating \(x\) with respect to \(\theta\), we get: $$\frac{dx}{d\theta} = 2\frac{d}{d\theta}\tan^{-1}\theta = 2\frac{1}{1+\theta^2}$$

Rewrite \(\sin x\) and \(\cos x\) in terms of \(\theta\)

Using the provided trigonometric identities, let's rewrite \(\sin x\) and \(\cos x\) in terms of \(\theta\): $$\sin x = 2\sin\frac{x}{2} \cos\frac{x}{2}$$ and $$\cos x = \cos^2\frac{x}{2}-\sin^2\frac{x}{2}$$ Now, let's find \(\sin\frac{x}{2}\) and \(\cos\frac{x}{2}\) using \(\tan\frac{x}{2}=\theta\): $$\sin\frac{x}{2} = \frac{\theta}{\sqrt{1+\theta^2}}$$ and $$\cos\frac{x}{2} = \frac{1}{\sqrt{1+\theta^2}}$$

Replace \(\sin x\), \(\cos x\), and \(dx\) in the integral

Now we can replace \(\sin x\), \(\cos x\), and \(dx\) in the integral with their expressions in terms of \(\theta\): $$\int \frac{dx}{1+\sin x+\cos x} = \int \frac{2 \frac{1}{1+\theta^2}}{1+2\frac{\theta}{\sqrt{1+\theta^2}}\frac{1}{\sqrt{1+\theta^2}}+\left(\frac{1}{\sqrt{1+\theta^2}}\right)^2-\left(\frac{\theta}{\sqrt{1+\theta^2}}\right)^2}d\theta$$

Simplify and solve the integral

The integral now can be simplified as: $$\int \frac{2\frac{1}{1+\theta^2}}{1+\frac{2\theta}{1+\theta^2}}d\theta$$ Now, let's multiply the numerator and denominator of the integrand by \((1+\theta^2)\): $$\int \frac{2}{(1+\theta^2)+2\theta}d\theta$$ Now, we have the integral in a more recognizable form: $$\int \frac{2}{(1+\theta^2)+2\theta}d\theta$$ To solve this integral, we can use the substitution method again with \(u = 1+\theta^2 + 2\theta\). Then, \(du = 2(1+\theta)d\theta\). So, the integral is: $$\int \frac{1}{u}du$$ Now, we can easily find the antiderivative: $$\int \frac{1}{u}du = \ln |u| + C = \ln |1+\theta^2 +2\theta |+C$$

Convert back to \(x\)

To convert the result back to \(x\), we need to express \(\theta\) in terms of \(x\). We know that \(x=2\tan^{-1}\theta\), so \(\tan\frac{x}{2}=\theta\). Now, we can rewrite the result: $$\ln |1+\theta^2 +2\theta |+C = \ln |1+(\tan\frac{x}{2})^2 +2\tan\frac{x}{2}|+C$$ So, the final answer is: $$\ln |1+(\tan\frac{x}{2})^2 +2\tan\frac{x}{2}|+C$$

Key concepts

Trigonometric Substitution

Trigonometric substitution is a powerful technique used to simplify integrals that involve square roots or specific algebraic expressions. By substituting trigonometric identities, complex expressions become manageable. In this problem, we used the substitution \( x = 2 \tan^{-1}\theta \) to transform the integral into a form that is easier to solve.
This method often involves converting a trigonometric expression into another using identities like \( \tan \), \( \sin \), and \( \cos \) functions. For this particular exercise, \( \sin x \), \( \cos x \), and \( dx \) were expressed in terms of \( \theta \), allowing us to manage the integral effectively.
This technique is especially useful when dealing with integrals involving quadratic terms, as it can linearize the expression and pave the way to straightforward integration.

Antiderivatives

Finding antiderivatives, or the process of integration, is like solving an inverse problem. It involves determining an original function given its derivative. In the given exercise, the goal was to evaluate an integral involving trigonometric identities by finding the antiderivative.
The key is to simplify the integrand and then integrate. Once the expression is simplified, finding the antiderivative becomes much simpler. For instance, after substitution and simplification, the integral became \( \int \frac{1}{u} du \).
This is a standard form integral, and its antiderivative is \( \ln|u| + C \), where \( C \) is a constant of integration. By recognizing these patterns, the integration process becomes more approachable, even when the initial integral seems complex.

Trigonometric Identities

Trigonometric identities are fundamental tools in manipulating and simplifying expressions in calculus, especially when dealing with integrals involving trigonometric functions. They help transform complex expressions into simpler forms that are easier to integrate.
In this exercise, the identities \( \sin x = 2 \sin\frac{x}{2} \cos\frac{x}{2} \) and \( \cos x = \cos^2\frac{x}{2} - \sin^2\frac{x}{2} \) were crucial. They allowed us to express \( \sin x \) and \( \cos x \) in terms of half-angle formulas, which were then related to \( \theta \). This made the integral more manageable.
Using trigonometric identities efficiently can dramatically simplify the process of solving integrals by reducing them to forms where standard integration techniques can be easily applied.