Question

Let $f(x) = \int \frac{\sin x + \cos x}{e^{-x} + \sin x} dx$ and $f(0)=1$, then $f(x)$ is equal to:

• A. $\ln(e^{-x}+\sin x)+1$
• B. $x+\ln(e^{-x}+\sin x)+1$
• C. $\ln(1+e^x \sin x)$
• D. $x+\ln(e^{x}+\sin x)+1$

Answer 1

Answer: (B)

x+ln(e^{-x}+sin x)+1

Answer 2

To evaluate the integral $f(x) = \int \frac{\sin x + \cos x}{e^{-x} + \sin x} dx$, we can manipulate the integrand.

The denominator is $e^{-x} + \sin x$. We can rewrite $e^{-x}$ as $\frac{1}{e^x}$. So, the denominator becomes $\frac{1}{e^x} + \sin x = \frac{1 + e^x \sin x}{e^x}$.

Substitute this back into the integral: $f(x) = \int \frac{\sin x + \cos x}{\frac{1 + e^x \sin x}{e^x}} dx$ $f(x) = \int \frac{e^x (\sin x + \cos x)}{1 + e^x \sin x} dx$

Now, this integral is in the form $\int \frac{g'(x)}{g(x)} dx$. Let $g(x) = 1 + e^x \sin x$. To find $g'(x)$, we differentiate $g(x)$ with respect to $x$: $g'(x) = \frac{d}{dx}(1 + e^x \sin x)$ Using the product rule for $e^x \sin x$: $\frac{d}{dx}(uv) = u'v + uv'$. Here, $u = e^x$ and $v = \sin x$. So $u' = e^x$ and $v' = \cos x$. $g'(x) = 0 + (e^x \sin x + e^x \cos x)$ $g'(x) = e^x (\sin x + \cos x)$

We can see that the numerator of our manipulated integrand is exactly $g'(x)$. So, the integral becomes: $f(x) = \int \frac{g'(x)}{g(x)} dx = \ln|g(x)| + C$ Substitute back $g(x) = 1 + e^x \sin x$: $f(x) = \ln|1 + e^x \sin x| + C$

Now, we use the given condition $f(0)=1$ to find the value of $C$. $f(0) = \ln|1 + e^0 \sin 0| + C$ Since $e^0 = 1$ and $\sin 0 = 0$: $f(0) = \ln|1 + 1 \cdot 0| + C$ $f(0) = \ln|1| + C$ $f(0) = 0 + C$ Given $f(0)=1$, so $C=1$.

Thus, the function $f(x)$ is: $f(x) = \ln|1 + e^x \sin x| + 1$

Now, let's compare this result with the given options. Notice that the options do not have absolute values. We should match the algebraic form. Let's look at Option 2: $x+\ln(e^{-x}+\sin x)+1$. Let's simplify the logarithmic term in Option 2: $\ln(e^{-x}+\sin x) = \ln\left(\frac{1}{e^x} + \sin x\right)$ $= \ln\left(\frac{1 + e^x \sin x}{e^x}\right)$ Using the logarithm property $\ln\left(\frac{A}{B}\right) = \ln A - \ln B$: $= \ln(1 + e^x \sin x) - \ln(e^x)$ $= \ln(1 + e^x \sin x) - x$ Substitute this back into Option 2: $f(x) = x + (\ln(1 + e^x \sin x) - x) + 1$ $f(x) = \ln(1 + e^x \sin x) + 1$ This matches our derived function $f(x)$, ignoring the absolute value sign which is common in multiple-choice questions when the domain is implicitly assumed to make the argument of logarithm positive.

The final answer is $x+\ln(e^{-x}+\sin x)+1$.