Question

Using one computer algebra system, it was found that \(\int \frac{d x}{1+\sin x}=\frac{\sin x-1}{\cos x},\) and using another computer algebra system, it was found that \(\int \frac{d x}{1+\sin x}=\frac{2 \sin (x / 2)}{\cos (x / 2)+\sin (x / 2)} \cdot\) Reconcile the two answers.

Short answer

Why or why not? Answer: Yes, the given computer algebra system answers reconcile because the derivatives of both given solutions match the original integrand function, \(\frac{1}{1+\sin{x}}\). The two solutions differ by a constant, which is an acceptable property for indefinite integrals.

Step-by-step solution

Differentiating the first solution

Differentiate the first given solution with respect to x: \(\frac{d}{dx} \left(\frac{\sin{x} - 1}{\cos{x}}\right)\) To differentiate a fraction, use the quotient rule: \((\frac{u}{v})' = \frac{u'v - uv'}{v^2}\) Here, \(u = \sin{x} - 1\) and \(v = \cos{x}\). Differentiate u and v with respect to x: \(u' = \frac{d}{dx}(\sin{x} - 1) = \cos{x}\) \(v' = \frac{d}{dx}(\cos{x}) = -\sin{x}\) Now, apply the quotient rule: \(\frac{d}{dx}\left(\frac{\sin{x} - 1}{\cos{x}}\right) = \frac{\cos{x} \cdot \cos{x} - (-\sin{x})\left(\sin{x} - 1\right)}{(\cos{x})^2}\) After simplifying, we get: \(\frac{d}{dx}\left(\frac{\sin{x} - 1}{\cos{x}}\right) = \frac{1}{1+\sin{x}}\)

Differentiating the second solution

Differentiate the second given solution with respect to x: \(\frac{d}{dx} \left(\frac{2\sin{(\frac{x}{2})}}{\cos{(\frac{x}{2})}+\sin{(\frac{x}{2})}}\right)\) Again, using the quotient rule with \(u = 2\sin{(\frac{x}{2})}\) and \(v = \cos{(\frac{x}{2})}+\sin{(\frac{x}{2})}\), we differentiate u and v with respect to x: \(u' = \frac{d}{dx}(2\sin{(\frac{x}{2})}) = \cos{(\frac{x}{2})}\) \(v' = \frac{d}{dx}(\cos{(\frac{x}{2})}+\sin{(\frac{x}{2})}) = -\frac{1}{2}\sin{(\frac{x}{2})} + \frac{1}{2}\cos{(\frac{x}{2})}\) Now applying the quotient rule: \(\frac{d}{dx}\left(\frac{2\sin{(\frac{x}{2})}}{\cos{(\frac{x}{2})}+\sin{(\frac{x}{2})}}\right) = \frac{\cos{(\frac{x}{2})}\left(\cos{(\frac{x}{2})}+\sin{(\frac{x}{2})}\right)- \left(-\frac{1}{2}\sin{(\frac{x}{2})} + \frac{1}{2}\cos{(\frac{x}{2})}\right)2\sin{(\frac{x}{2})}}{(\cos{(\frac{x}{2})}+\sin{(\frac{x}{2})})^2}\) After simplifying, we get: \(\frac{d}{dx}\left(\frac{2\sin{(\frac{x}{2})}}{\cos{(\frac{x}{2})}+\sin{(\frac{x}{2})}}\right) = \frac{1}{1+\sin{x}}\)

Conclusion

Since the derivatives of both given solutions match the original integrand function, \(\frac{1}{1+\sin{x}}\), the two solutions differ by a constant. This is an acceptable property for indefinite integrals, and the given computer algebra system answers can be reconciled.

Key concepts

Quotient Rule

When you have a function that is the ratio of two other functions, the quotient rule is an essential tool for differentiation. This comes in handy especially for problems that involve complex fractions. Suppose you have a function in the form of \(\frac{u}{v}\), where both \(u\) and \(v\) are differentiable functions of \(x\).

The quotient rule states that the derivative of \(\frac{u}{v}\) with respect to \(x\) is:

• \((\frac{u}{v})' = \frac{u'v - uv'}{v^2}\)

This means that you find the derivatives \(u'\) and \(v'\) first.
Then, you plug them into the formula to compute the derivative for the whole fraction. In our exercise, this principle was applied effectively to differentiate each of the given integral solutions by treating the numerator and denominator separately.

Always remember: the quotient rule is a cornerstone in calculus that makes dealing with division of functions much simpler!

Differentiation

Differentiation is the process of finding the derivative of a function. The derivative represents the rate of change, or slope, of the function at any given point. This is crucial for understanding how functions behave.

In our exercise, differentiation was used to verify that two different expressions both have the same derivative, which is equivalent to the integrand \(\frac{1}{1+\sin{x}}\). By differentiating these expressions and comparing them to the original integrand, we confirmed that the differences in the integrals only stemmed from a constant.

There are many rules in differentiation, including the power rule, product rule, chain rule, and of course, the quotient rule. All of these help to systematically break down and solve equations involving derivatives. The key is to know which rule to apply for a given function form.

Indefinite Integrals

Indefinite integrals represent a family of functions that have been integrated without specifying the limits of integration.
A key point to note is that indefinite integrals include an arbitrary constant \(C\), symbolizing that there are infinitely many antiderivatives.

In our exercise, the task was to reconcile two different indefinite integrals that yielded different forms but the same derivative. This highlights a fundamental property: indefinite integrals that differ by a constant still represent the same family of antiderivatives.

• Example from our exercise: \(\int \frac{dx}{1+\sin x}\)
• Solutions: \(\frac{\sin x - 1}{\cos x}\) and \(\frac{2 \sin{(x/2)}}{\cos{(x/2)} + \sin{(x/2)}}\)

By verifying that both have the same derivative, we see they differ only by a constant, which is permissible due to the nature of indefinite integrals.

Remember: When dealing with indefinite integrals, one must always consider the constant \(C\), as it captures the infinite possibilities of antiderivatives for a function.