Question

If $f(x) = \int \frac{d(\sin x - \cos x)}{1-\sin 2x}, f(0) = 1$, then the function $f(x)$ is

• A. $\frac{1}{\sqrt{2}}cosec(x+\frac{\pi}{4})$
• B. $\frac{1}{\cos x + \sin x}$
• C. $\frac{1}{\sqrt{2}}sec(x+\frac{\pi}{4})$
• D. $1-\sin 2x$

Answer 1

Answer: (C)

$\frac{1}{\sqrt{2}}sec(x+\frac{\pi}{4})$

Answer 2

The problem asks us to find the function $f(x)$ given its integral form and an initial condition.

1. Simplify the integrand: The given integral is $f(x) = \int \frac{d(\sin x - \cos x)}{1-\sin 2x}$.

First, let's analyze the numerator: $d(\sin x - \cos x) = (\frac{d}{dx}(\sin x - \cos x)) dx = (\cos x - (-\sin x)) dx = (\cos x + \sin x) dx$.

Next, let's simplify the denominator using trigonometric identities: We know that $1 = \sin^2 x + \cos^2 x$ and $\sin 2x = 2\sin x \cos x$. So, $1 - \sin 2x = \sin^2 x + \cos^2 x - 2\sin x \cos x$. This is the expansion of a perfect square: $(\sin x - \cos x)^2$. Thus, $1 - \sin 2x = (\sin x - \cos x)^2$.

Substituting these simplified forms back into the integral, we get: $f(x) = \int \frac{(\cos x + \sin x) dx}{(\sin x - \cos x)^2}$.

2. Perform substitution to solve the integral: Let $u = \sin x - \cos x$. Then, the differential $du = (\cos x + \sin x) dx$.

Substituting $u$ and $du$ into the integral: $f(x) = \int \frac{du}{u^2} = \int u^{-2} du$.

Now, integrate $u^{-2}$ with respect to $u$: $f(x) = \frac{u^{-2+1}}{-2+1} + C = \frac{u^{-1}}{-1} + C = -\frac{1}{u} + C$.

Substitute back $u = \sin x - \cos x$: $f(x) = -\frac{1}{\sin x - \cos x} + C = \frac{1}{\cos x - \sin x} + C$.

3. Use the initial condition to find the constant of integration (C): We are given that $f(0) = 1$. Substitute $x=0$ into the expression for $f(x)$: $f(0) = \frac{1}{\cos 0 - \sin 0} + C$. We know that $\cos 0 = 1$ and $\sin 0 = 0$. So, $f(0) = \frac{1}{1 - 0} + C = 1 + C$.

Given $f(0) = 1$, we have: $1 + C = 1$ This implies $C = 0$.

Therefore, the function $f(x)$ is: $f(x) = \frac{1}{\cos x - \sin x}$.

4. Compare with the given options: Let's simplify each option to see which one matches our derived $f(x)$.

• A. $\frac{1}{\sqrt{2}}\csc(x+\frac{\pi}{4})$ $\csc(x+\frac{\pi}{4}) = \frac{1}{\sin(x+\frac{\pi}{4})}$. Using the sine addition formula $\sin(A+B) = \sin A \cos B + \cos A \sin B$: $\sin(x+\frac{\pi}{4}) = \sin x \cos \frac{\pi}{4} + \cos x \sin \frac{\pi}{4} = \sin x \cdot \frac{1}{\sqrt{2}} + \cos x \cdot \frac{1}{\sqrt{2}} = \frac{1}{\sqrt{2}}(\sin x + \cos x)$. So, Option A becomes $\frac{1}{\sqrt{2} \cdot \frac{1}{\sqrt{2}}(\sin x + \cos x)} = \frac{1}{\sin x + \cos x}$. This does not match $f(x)$.

• B. $\frac{1}{\cos x + \sin x}$ This is the same as the simplified form of Option A, so it does not match $f(x)$.

• C. $\frac{1}{\sqrt{2}}\sec(x+\frac{\pi}{4})$ $\sec(x+\frac{\pi}{4}) = \frac{1}{\cos(x+\frac{\pi}{4})}$. Using the cosine addition formula $\cos(A+B) = \cos A \cos B - \sin A \sin B$: $\cos(x+\frac{\pi}{4}) = \cos x \cos \frac{\pi}{4} - \sin x \sin \frac{\pi}{4} = \cos x \cdot \frac{1}{\sqrt{2}} - \sin x \cdot \frac{1}{\sqrt{2}} = \frac{1}{\sqrt{2}}(\cos x - \sin x)$. So, Option C becomes $\frac{1}{\sqrt{2} \cdot \frac{1}{\sqrt{2}}(\cos x - \sin x)} = \frac{1}{\cos x - \sin x}$. This matches our derived $f(x)$.

• D. $1-\sin 2x$ This is clearly not $f(x)$.

Thus, option C is the correct answer.