Question

If \[ \int \frac{\sin\left(x-\frac{\pi}{4}\right)}{2+\sin 2x}\,dx = -\frac{1}{\sqrt2}\tan^{-1}(f(x))+C, \] then \(f(x)=\)

• A. \(\sin x-\cos x\)
• B. \(\sqrt2\cos\left(x-\frac{\pi}{4}\right)\)
• C. \(\sin\left(x-\frac{\pi}{4}\right)\)
• D. \(\sqrt2\tan\left(x-\frac{\pi}{4}\right)\)

Hint:

When the denominator contains \(2+\sin 2x\), try rewriting it using \[ 2+\sin 2x=1+(\sin x+\cos x)^2. \]

Answer 1

The Correct Option is B

Step 1: Rewrite the numerator and denominator.
\(\sin(x-\pi/4)=\dfrac{\sin x-\cos x}{\sqrt{2}}\). Let \(u = \sin x+\cos x = \sqrt{2}\cos(x-\pi/4)\). Then \(du=(\cos x-\sin x)dx = -(\sin x-\cos x)dx\).

Step 2: Rewrite the denominator.
\(2+\sin 2x = 1+(\sin x+\cos x)^2 = 1+u^2\).

Step 3: Integrate.
\[\int\frac{(\sin x-\cos x)/\sqrt{2}}{1+u^2}\,dx = \frac{1}{\sqrt{2}}\int\frac{-du}{1+u^2} = -\frac{1}{\sqrt{2}}\arctan(u)+C\] So \(f(x)=u=\sin x+\cos x = \sqrt{2}\cos(x-\pi/4)\).
\[ \boxed{f(x)=\sqrt{2}\cos(x-\pi/4)} \]