A.O.D and integration is all you need

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Calculation of a :

$\large \displaystyle \ln(1 + \sin(x)) = 0$

$\large \displaystyle 1 + \sin(x) = 1$

$\large \displaystyle \sin(x) = 0$

In the given interval:

$\large \displaystyle x = -2\pi, x= -\pi, x = 0, x = \pi, x = 2\pi$

Actually, since $\sin(x)$ is odd, we don't even have to calculate the zeroes to know that their sum is $0$ . So:

$\color{#20A900} \boxed{\large \displaystyle a = 0}$

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Calculation of b :

$\large \displaystyle f'(x) = \frac{\cos(x)}{1 + \sin(x)}$

$\large \displaystyle f''(x) = - \frac{1 + \sin(x)}{[1 + \sin(x)]^2}$

$\large \displaystyle f''(x) = - \frac{1}{1 + \sin(x)}$

Inflection points are points where $f''(x) =0$ . By $f''(x)$ , above, there aren't any. So:

$\color{#20A900} \boxed{\large \displaystyle b = 0}$

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Calculation of c :

$\large \displaystyle f'(x) = \frac{\cos(x)}{1 + \sin(x)} = 0$

$\large \displaystyle \cos(x) = 0$ while $\large \displaystyle 1 + \sin(x) \neq 0$

In the given interval:

$\large \displaystyle x = \frac{\pi}{2}, x = - \frac{3\pi}{2}$

So:

$\color{#20A900} \boxed{\large \displaystyle c = 2}$

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Calculation of d :

Since $1 + \sin(x)$ is bounded between $0$ and $2$ , the only way to make $f(x)$ go to $\infty$ or $-\infty$ for finite values of $x$ (which is the definition of an asymptote) is to make:

$\large \displaystyle 1 + \sin(x) = 0$

Which will make the argument of $\ln$ go to $0$ , i.e., $f(x)$ go to $-\infty$

$\large \displaystyle \sin(x) = -1$

In the given interval:

$\large \displaystyle x = -\frac{\pi}{2}, x = \frac{3\pi}{2}$

So:

$\color{#20A900} \boxed{\large \displaystyle d = 2}$

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Calculation of e :

$\large \displaystyle e = \frac{1}{\pi} \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \ln(1 + \sin(x)) dx$

$\large \displaystyle e = \frac{1}{\pi} \left [ \frac{1}{2} \left ( \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \ln(1 + \sin(x)) dx + \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \ln(1 + \sin(\pi/2 - \pi/2 - x)) dx \right ) \right ]$

$\large \displaystyle e = \frac{1}{\pi} \left [ \frac{1}{2} \left ( \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \ln(1 + \sin(x)) dx + \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \ln(1 - \sin(x)) dx \right ) \right ]$

$\large \displaystyle e = \frac{1}{\pi} \left [ \frac{1}{2} \left ( \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \ln[(1 + \sin(x))\cdot(1 - \sin(x))] dx \right ) \right ]$

$\large \displaystyle e = \frac{1}{\pi} \left [ \frac{1}{2} \left ( \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \ln(\cos^2(x)) dx \right ) \right ]$

$\large \displaystyle e = \frac{1}{\pi} \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \ln(\cos(x)) dx$

Since $\ln(\cos(x))$ is even:

$\large \displaystyle e = \frac{2}{\pi} \int_{0}^{\frac{\pi}{2}} \ln(\cos(x)) dx$

Make $u = \frac{\pi}{2} - x$ :

$\large \displaystyle e = \frac{2}{\pi} \int_{0}^{\frac{\pi}{2}} \ln(\sin(u)) du$

$\large \displaystyle e = \frac{2}{\pi} \int_{0}^{\frac{\pi}{2}} \ln(2 \sin(u/2) \cos(u/2)) du$

$\large \displaystyle e = \frac{2}{\pi} \left [ \frac{\pi \ln(2)}{2} + \int_{0}^{\frac{\pi}{2}} \ln(\sin(u/2))du + \int_{0}^{\frac{\pi}{2}} \ln(\cos(u/2))du \right ]$

Make $t = u/2$ on both integrals:

$\large \displaystyle e = \frac{2}{\pi} \left [ \frac{\pi \ln(2)}{2} + 2 \int_{0}^{\frac{\pi}{4}} \ln(\sin(t))dt + 2 \int_{0}^{\frac{\pi}{4}} \ln(\cos(t))dt \right ]$

On first integral, make $v = \pi/2 - t$ :

$\large \displaystyle e = \frac{2}{\pi} \left [ \frac{\pi \ln(2)}{2} - 2 \int_{\frac{\pi}{2}}^{\frac{\pi}{4}} \ln(\cos(\pi/2 - v))dv + 2 \int_{0}^{\frac{\pi}{4}} \ln(\cos(t))dt \right ]$

$\large \displaystyle e = \frac{2}{\pi} \left [ \frac{\pi \ln(2)}{2} + 2 \int_{\frac{\pi}{4}}^{\frac{\pi}{2}} \ln(\sin(v))dv + 2 \int_{0}^{\frac{\pi}{4}} \ln(\cos(t))dt \right ]$

$\large \displaystyle e = \frac{2}{\pi} \left [ \frac{\pi \ln(2)}{2} + 2 \int_{0}^{\frac{\pi}{2}} \ln(\sin(x))dx \right ]$

$\large \displaystyle e = \ln(2) + \frac{4}{\pi} \int_{0}^{\frac{\pi}{2}} \ln(\sin(x))dx$

$\large \displaystyle e = \ln(2) + 2e$

$\color{#20A900} \boxed{\large \displaystyle e = -\ln(2)}$

Thus:

$\color{#3D99F6} \boxed{\large \displaystyle a + b + c + d + e = 4 - \ln(2) \approx 3.307}$

• zeroes of f(x) can be easily seen as when sinx=0..hence sum of zeroes = 0.
• there aren't any inflection points for f(x),
• maxima at x = $\frac{\pi}{2}$ and $\frac{-3\pi}{2}$ and no minima
• asymptotes at x = $\frac{-\pi}{2}$ and $\frac{3\pi}{2}$
• def. integral has value = $\frac{-\pi*ln4}{2}$