Question

$\int_{0}^{\pi} \frac{x \cos x}{(1 + \sin x)^2} dx$

Answer 1

2 - \pi

Answer 2

To evaluate the integral $I = \int_{0}^{\pi} \frac{x \cos x}{(1 + \sin x)^2} dx$, we use integration by parts. The formula for integration by parts is $\int u \, dv = uv - \int v \, du$.

Let $u = x$ and $dv = \frac{\cos x}{(1 + \sin x)^2} dx$.

First, find $du$ and $v$: $du = dx$.

To find $v$, integrate $dv$: $v = \int \frac{\cos x}{(1 + \sin x)^2} dx$. Let $t = 1 + \sin x$. Then $dt = \cos x \, dx$. So, $v = \int \frac{dt}{t^2} = -\frac{1}{t} = -\frac{1}{1 + \sin x}$.

Now, substitute these into the integration by parts formula: $I = \left[ x \left(-\frac{1}{1 + \sin x}\right) \right]_{0}^{\pi} - \int_{0}^{\pi} \left(-\frac{1}{1 + \sin x}\right) dx$ $I = \left[ -\frac{x}{1 + \sin x} \right]_{0}^{\pi} + \int_{0}^{\pi} \frac{1}{1 + \sin x} dx$

Evaluate the first term: $\left[ -\frac{x}{1 + \sin x} \right]_{0}^{\pi} = \left(-\frac{\pi}{1 + \sin \pi}\right) - \left(-\frac{0}{1 + \sin 0}\right)$ Since $\sin \pi = 0$ and $\sin 0 = 0$: $= \left(-\frac{\pi}{1 + 0}\right) - \left(-\frac{0}{1 + 0}\right)$ $= -\pi - 0 = -\pi$.

Now, we need to evaluate the second integral: $J = \int_{0}^{\pi} \frac{1}{1 + \sin x} dx$. We can simplify the integrand by multiplying the numerator and denominator by $(1 - \sin x)$: $J = \int_{0}^{\pi} \frac{1 - \sin x}{(1 + \sin x)(1 - \sin x)} dx = \int_{0}^{\pi} \frac{1 - \sin x}{1 - \sin^2 x} dx$ $J = \int_{0}^{\pi} \frac{1 - \sin x}{\cos^2 x} dx = \int_{0}^{\pi} \left( \frac{1}{\cos^2 x} - \frac{\sin x}{\cos^2 x} \right) dx$ $J = \int_{0}^{\pi} (\sec^2 x - \sec x \tan x) dx$

The antiderivative of $\sec^2 x - \sec x \tan x$ is $\tan x - \sec x$. However, $\tan x$ and $\sec x$ are undefined at $x = \pi/2$, which is within the integration interval $[0, \pi]$. Thus, we must evaluate the integral as an improper integral or use a property of definite integrals.

Using the property $\int_0^{2a} f(x) dx = 2 \int_0^a f(x) dx$ if $f(2a-x) = f(x)$. Let $f(x) = \frac{1}{1 + \sin x}$. Here $2a = \pi$, so $a = \pi/2$. $f(\pi - x) = \frac{1}{1 + \sin(\pi - x)} = \frac{1}{1 + \sin x} = f(x)$. So, $J = \int_{0}^{\pi} \frac{1}{1 + \sin x} dx = 2 \int_{0}^{\pi/2} \frac{1}{1 + \sin x} dx$.

Now, we evaluate $J' = \int_{0}^{\pi/2} (\sec^2 x - \sec x \tan x) dx$. $J' = [\tan x - \sec x]_{0}^{\pi/2}$. Since the antiderivative is undefined at $\pi/2$, we take the limit: $J' = \lim_{b \to \pi/2^-} (\tan b - \sec b) - (\tan 0 - \sec 0)$ $J' = \lim_{b \to \pi/2^-} \left( \frac{\sin b}{\cos b} - \frac{1}{\cos b} \right) - (0 - 1)$ $J' = \lim_{b \to \pi/2^-} \left( \frac{\sin b - 1}{\cos b} \right) + 1$. This limit is of the form $0/0$, so we apply L'Hopital's rule: $J' = \lim_{b \to \pi/2^-} \left( \frac{\cos b}{-\sin b} \right) + 1$ $J' = \frac{\cos(\pi/2)}{-\sin(\pi/2)} + 1 = \frac{0}{-1} + 1 = 0 + 1 = 1$.

So, $\int_{0}^{\pi/2} \frac{1}{1 + \sin x} dx = 1$. Therefore, $J = 2 \times 1 = 2$.

Substitute this value back into the expression for $I$: $I = -\pi + J$ $I = -\pi + 2$.

The final answer is $2 - \pi$.