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: Simplify the numerator.
We know that \[ \sin\left(x-\frac{\pi}{4}\right) = \frac{\sin x-\cos x}{\sqrt2}. \] So, \[ \int \frac{\sin\left(x-\frac{\pi}{4}\right)}{2+\sin 2x}\,dx = \frac{1}{\sqrt2} \int \frac{\sin x-\cos x}{2+\sin 2x}\,dx. \]

Step 2: Simplify the denominator.
Since \[ \sin 2x=2\sin x\cos x, \] we have \[ 2+\sin 2x=2+2\sin x\cos x. \] Also, \[ (\sin x+\cos x)^2=1+2\sin x\cos x. \] Therefore, \[ 2+\sin 2x=1+(\sin x+\cos x)^2. \]

Step 3: Use substitution.
Let \[ u=\sin x+\cos x. \] Then, \[ \frac{du}{dx}=\cos x-\sin x. \] So, \[ du=(\cos x-\sin x)\,dx. \] Hence, \[ (\sin x-\cos x)\,dx=-du. \]

Step 4: Transform the integral.
\[ \frac{1}{\sqrt2} \int \frac{\sin x-\cos x}{1+(\sin x+\cos x)^2}\,dx = -\frac{1}{\sqrt2} \int \frac{du}{1+u^2}. \] \[ = -\frac{1}{\sqrt2}\tan^{-1}u+C. \] Substituting back, \[ = -\frac{1}{\sqrt2}\tan^{-1}(\sin x+\cos x)+C. \]

Step 5: Identify \(f(x)\).
Comparing with \[ -\frac{1}{\sqrt2}\tan^{-1}(f(x))+C, \] we get \[ f(x)=\sin x+\cos x. \] Now, \[ \sqrt2\cos\left(x-\frac{\pi}{4}\right) = \sqrt2\left(\frac{\cos x+\sin x}{\sqrt2}\right) = \sin x+\cos x. \] Thus, \[ f(x)=\sqrt2\cos\left(x-\frac{\pi}{4}\right). \]

Step 6: Final conclusion.
Therefore, \[ \boxed{\sqrt2\cos\left(x-\frac{\pi}{4}\right)} \]