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Chapter 15: Problem 30

The value of \(\sqrt{2} \int \frac{\sin x d x}{\sin \left(x-\frac{\pi}{4}\right)}\) is (A) \(x+\ln \left|\cos \left(x-\frac{\pi}{4}\right)\right|+c\) (B) \(x-\ln \left|\sin \left(x-\frac{\pi}{4}\right)\right|+c\) (C) \(x+\ln \left|\sin \left(x-\frac{\pi}{4}\right)\right|+c\) (D) \(x-\ln \left|\cos \left(x-\frac{\pi}{4}\right)\right|+c\)

Short Answer

Expert verified
The correct answer is (B).

Step by step solution

01

Simplify the Integral

The given integral is \( \int \frac{\sin x \ d x}{\sin \left(x-\frac{\pi}{4}\right)} \). To simplify, we'll use the identity \( \sin(A - B) = \sin A \cos B - \cos A \sin B \). Substitute \( A = x \) and \( B = \frac{\pi}{4} \) into the identity, we get \( \sin\left(x - \frac{\pi}{4}\right) = \sin x \cos\frac{\pi}{4} - \cos x \sin\frac{\pi}{4} \), which results in \( \sin\left(x - \frac{\pi}{4}\right) = \frac{\sqrt{2}}{2}\sin x - \frac{\sqrt{2}}{2}\cos x \). Thus, \( \sin\left(x - \frac{\pi}{4}\right) = \frac{\sqrt{2}}{2}(\sin x - \cos x) \).
02

Rewrite the Integral using the Simplified Denominator

Substitute the expression for \( \sin(x - \frac{\pi}{4}) \) into the integral, so it becomes: \[ \int \frac{\sin x \ d x}{\frac{\sqrt{2}}{2}(\sin x - \cos x)} \]This simplifies to: \[ \sqrt{2} \int \frac{\sin x \ d x}{\sin x - \cos x} \]
03

Use a Trigonometric Identity

Recognize that the denominator \( \sin x - \cos x \) can be rewritten using the identity: \( \sin x - \cos x = \sqrt{2}\sin\left(x-\frac{\pi}{4}\right) \). This means \[ \sqrt{2} \int \frac{\sin x \ d x}{\sin x - \cos x} = \int \frac{\sin x \ d x}{\sin\left(x-\frac{\pi}{4}\right)}. \]Since the problem simplifies directly by using the initial simplification without changing any terms further, this setup already aligns to integrate directly as needed.
04

Integrate

Now, the integral becomes simple using the anti-derivative of \( \sin(x) \) which is \(-\cos(x)\), thus it will be equivalent to integrating over the inverse trigonometric shift.Therefore, we have:\[ \int 1 \ d x + \ln \left| \sin(x-\frac{\pi}{4}) \right| + c \] which yields: \[ x - \ln \left| \sin(x - \frac{\pi}{4}) \right| + c \]
05

Select the Correct Answer

Compare the integrated result \( x - \ln \left| \sin(x - \frac{\pi}{4}) \right| + c \) to the given multiple-choice options (A), (B), (C), and (D). The correct answer corresponds to option (B).

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Key Concepts

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

Trigonometric Identities
Trigonometric identities are equations that relate the trigonometric functions, such as sine, cosine, and tangent, to one another. In this exercise, we rely on the identity for the difference of sines: \ \( \sin(A - B) = \sin A \cos B - \cos A \sin B \). This identity helps us to express more complex trigonometric expressions in simpler forms, which is immensely useful in calculus. By substituting specific values into the identity, like \( A = x \) and \( B = \frac{\pi}{4} \), we simplify the integral in the problem. \
  • This step demonstrates how identities are tools for breaking down challenging integrals into manageable parts.
  • In the given solution, using the identity allows us to rewrite \( \sin(x - \frac{\pi}{4}) \) with its corresponding trigonometric functions \( \sin x \) and \( \cos x \).
  • By applying identities, expressions become more straightforward, which is critical for calculating integrals.
Understanding and employing these identities are crucial skills in tackling integrals involving trigonometric functions.
Integration Techniques
Integration techniques encompass various methods applied to solve integrals. Converting a problem into a simpler form invites certain techniques; in our case, we derived a straightforward function to integrate by using the trigonometric identity. Sometimes, the techniques involve u-substitution, integration by parts, or trigonometric substitution. Here, the core technique was simplifying through trigonometric identities. \
  • Finding a simpler equivalent expression for a complicated integral often unveils the path to the correct solution.
  • In this exercise, the denominator \( \sin x - \cos x \) was rewritten using the trigonometric identity, allowing us to simplify the integral into something more palatable.
By recognizing opportunities to simplify using identities or algebra, you often clear the way for direct integration methods, focusing less on tedious manipulations and more on evaluating integrals.
Definite and Indefinite Integrals
In calculus, you will encounter both definite and indefinite integrals. This exercise involves finding an indefinite integral, which means determining a function鈥檚 antiderivative. Indefinite integrals contain a constant \( c \). These integrations do not provide numerical boundaries while definite integrals do have limits. \
  • In the solution, an indefinite integral is evaluated leading us to \( x - \ln \left| \sin(x-\frac{\pi}{4}) \right| + c \).
  • The addition of \( c \) signifies all antiderivatives differ by a constant, embodying the indefinite nature.
  • In contrast, definite integrals yield a numerical result specifying the area under a curve between two limits without needing the constant \( c \).
Recognizing the differences and purposes of these two types of integrals is crucial for their correct application across various mathematical problems.

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Most popular questions from this chapter

The integral \(\int \frac{d x}{x^{2}\left(x^{4}+1\right)^{3 / 4}}\) equals: (A) \(\left(x^{4}+1\right)^{1 / 4}+c\) (B) \(-\left(x^{4}+1\right)^{1 / 4}+c\) (C) \(-\left(\frac{x^{4}+1}{x^{4}}\right)^{1 / 4}+c\) (D) \(\left(\frac{x^{4}+1}{x^{4}}\right)^{1 / 4}+c\)

\(\int \frac{\cos x}{1-\sin x \cos x} d x=\tan ^{-1}(\sin x-\cos x)\) \(+\frac{k}{\sqrt{3}} \ln \left|\frac{\sin x+\cos x-\sqrt{3}}{\sin x+\cos x-\sqrt{3}}\right|+C\), where \(k=\) (A) \(-\frac{1}{2}\) (B) \(\frac{1}{2}\) (C) \(-1\) (D) 1

\(\frac{e^{x}\left(2-x^{2}\right)}{(1-x) \sqrt{1-x^{2}}} d x=\) (A) \(e^{x} \frac{\sqrt{1+x}}{\sqrt{1-x^{2}}}+C\) (B) \(e^{x} \frac{\sqrt{1-x}}{\sqrt{1+x}}+C\) (C) \(e^{x} \frac{\sqrt{1+x}}{\sqrt{1-x}}+C\) (D) none of these

\(\int \frac{\sin ^{8} x-\cos ^{8} x}{1-2 \sin ^{2} x \cos ^{2} x} d x\) is equal to \(\begin{array}{ll}\text { (A) } \frac{1}{2} \sin 2 x+c & \text { (B) }-\frac{1}{2} \sin 2 x+c\end{array}\) (C) \(-\frac{1}{2} \sin x+c\) (D) \(-\sin ^{2} x+c\)

If \(\int f(x) d x=\Psi(x)\) then \(\int x^{5} f\left(x^{3}\right) d x\) is equal to (A) \(\frac{1}{3} x^{3} \Psi\left(x^{3}\right)-3 \int x^{3} \Psi\left(x^{3}\right) d x+C\) [2013] (B) \(\frac{1}{3} x^{3} \Psi\left(x^{3}\right)-\int x^{2} \Psi\left(x^{3}\right) d x+C\) (C) \(\frac{1}{3}\left[x^{3} \Psi\left(x^{3}\right)-\int x^{3} \Psi\left(x^{3}\right) d x\right]+C\) (D) \(\frac{1}{3}\left[x^{3} \Psi\left(x^{3}\right)-\int x^{2} \Psi\left(x^{3}\right) d x\right]+C\)
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