Question

If \( \int \frac{\sin(x-\pi/4)}{2+\sin 2x} dx = -\frac{1}{\sqrt{2}} \tan^{-1}(f(x)) + C \), then \(f(x) = \)

• A. \( \sin x - \cos x \)
• B. \( \sqrt{2}\cos(x-\pi/4) \)
• C. \( \sin(x-\pi/4) \)
• D. \( \sqrt{2}\tan(x-\pi/4) \)

Hint:

Expand \(\sin(A-B)\).
Use \(1+\sin 2x = (\sin x + \cos x)^2\) or \(2+\sin 2x = 1 + (\sin x + \cos x)^2\).
Use substitution \(u = \sin x + \cos x\), then \(du = (\cos x - \sin x)dx\).
\(\int \frac{1}{1+u^2}du = \tan^{-1}u + C\).

Answer 1

The Correct Option is A

Let \(I = \int \frac{\sin(x-\pi/4)}{2+\sin 2x} dx\). Numerator: \( \sin(x-\pi/4) = \sin x \cos(\pi/4) - \cos x \sin(\pi/4) = \frac{1}{\sqrt{2}}(\sin x - \cos x) \). Denominator: \(2+\sin 2x\). We know \( (\sin x + \cos x)^2 = \sin^2 x + \cos^2 x + 2\sin x \cos x = 1 + \sin 2x \). So, \( \sin 2x = (\sin x + \cos x)^2 - 1 \). Then \(2+\sin 2x = 2 + (\sin x + \cos x)^2 - 1 = 1 + (\sin x + \cos x)^2 \). So the integral becomes: \[ I = \int \frac{\frac{1}{\sqrt{2}}(\sin x - \cos x)}{1 + (\sin x + \cos x)^2} dx \] Let \(u = \sin x + \cos x\). Then \(du = (\cos x - \sin x)dx = -(\sin x - \cos x)dx\). So, \((\sin x - \cos x)dx = -du\). Substitute these into the integral: \[ I = \int \frac{\frac{1}{\sqrt{2}}(-du)}{1+u^2} = -\frac{1}{\sqrt{2}} \int \frac{du}{1+u^2} \] \[ I = -\frac{1}{\sqrt{2}} \tan^{-1}(u) + C \] Substitute back \(u = \sin x + \cos x\): \[ I = -\frac{1}{\sqrt{2}} \tan^{-1}(\sin x + \cos x) + C \] The problem states \( I = -\frac{1}{\sqrt{2}} \tan^{-1}(f(x)) + C \). Comparing, we have \(f(x) = \sin x + \cos x\). This is not directly option (a). Option (a) is \(\sin x - \cos x\). Let's check if there's a sign error in my \(du\). \(u = \sin x + \cos x \implies du = (\cos x - \sin x)dx\). Numerator is \( (\sin x - \cos x)dx \). So \( (\sin x - \cos x)dx = -du \). This is correct. The form of \(f(x)\) I got is \(\sin x + \cos x\). Let's re-examine the options. The checkmark is on (a) \( \sin x - \cos x \). If \(f(x) = \sin x - \cos x\), then \(u\) in my substitution should have been \( \sin x - \cos x \). Let \(u = \sin x - \cos x\). Then \(du = (\cos x + \sin x)dx\). Denominator: \(1+(\sin x + \cos x)^2\). Numerator: \(\frac{1}{\sqrt{2}}(\sin x - \cos x)dx\). This substitution does not directly simplify the denominator in terms of u. Let's check the options again from the image. 1. \(\sin x - \cos x\) 2. \(\sqrt{2}\cos(x-\pi/4)\) 3. \(\sin(x-\pi/4)\) 4. \(\sqrt{2}\tan(x-\pi/4)\) The checkmark is on option (a) \(\sin x - \cos x\). My derived \(f(x) = \sin x + \cos x\). Is \(\sin x + \cos x\) related to \( \sin x - \cos x \) or other options? \(\sin x + \cos x = \sqrt{2}(\frac{1}{\sqrt{2}}\sin x + \frac{1}{\sqrt{2}}\cos x) = \sqrt{2}(\cos(\pi/4)\sin x + \sin(\pi/4)\cos x) = \sqrt{2}\sin(x+\pi/4)\). Or \(\sqrt{2}(\sin(\pi/4)\sin x + \cos(\pi/4)\cos x) = \sqrt{2}\cos(x-\pi/4)\). So, \(\sin x + \cos x = \sqrt{2}\cos(x-\pi/4)\). This is option (b). If my \(f(x) = \sin x + \cos x\) is correct, then option (b) should be the answer. The checkmark is on (a). Where could the difference arise? The problem might be set up so that the derivative of \(f(x)\) appears in the numerator (scaled). If \(f(x) = \sin x - \cos x\), then \(f'(x) = \cos x - (-\sin x) = \cos x + \sin x\). The numerator of the integrand has \(\sin x - \cos x\). The denominator is \(1+(\sin x + \cos x)^2\). If the form is \( \int \frac{g'(x)}{1+(g(x))^2} dx \), then it's \(\tan^{-1}(g(x))\). Let \(g(x) = \sin x + \cos x\). Then \(g'(x) = \cos x - \sin x\). Numerator is \(\frac{1}{\sqrt{2}}(\sin x - \cos x) = -\frac{1}{\sqrt{2}}(\cos x - \sin x) = -\frac{1}{\sqrt{2}}g'(x)\). So \(I = \int -\frac{1}{\sqrt{2}} \frac{g'(x)}{1+(g(x))^2} dx = -\frac{1}{\sqrt{2}} \tan^{-1}(g(x)) + C\). \(I = -\frac{1}{\sqrt{2}} \tan^{-1}(\sin x + \cos x) + C\). So, \(f(x) = \sin x + \cos x\). This can be written as \(\sqrt{2}\cos(x-\pi/4)\) which is option (b). The provided answer key (checkmark on (a)) suggests \(f(x) = \sin x - \cos x\). This would require the integral to be \(-\frac{1}{\sqrt{2}}\tan^{-1}(\sin x - \cos x)\). For this, the numerator (after extracting \(-1/\sqrt{2}\)) should be the derivative of \(\sin x - \cos x\), which is \(\cos x + \sin x\). The actual numerator is \( \sin x - \cos x \). So the \(f(x)\) from my derivation is \( \sin x + \cos x \equiv \sqrt{2}\cos(x-\pi/4) \). The marked answer (a) is different. There is a discrepancy. I will stick to my derivation. The closest matching option to my derived \(f(x)\) is (b). \[ \boxed{\sin x + \cos x \text{ (which is } \sqrt{2}\cos(x-\pi/4) \text{, option (b). Option (a) seems incorrect.)}} \]