`int_(0)^(pi//2)(cos x)/((1+sin x)^(3))dx=`

Options

A. `(8)/(3)`
B. `(3)/(8)`
C. `(33)/(88)`
D. `(88)/(33)`

Answer

To solve the integral

\[ I = \int_{0}^{\frac{\pi}{2}} \frac{\cos x}{(1 + \sin x)^{3}} \, dx, \]

we will use a substitution method.

### Step 1: Substitution
Let \( t = 1 + \sin x \). Then, we differentiate both sides to find \( dx \):
\[ dt = \cos x \, dx \implies dx = \frac{dt}{\cos x}. \]

### Step 2: Change Limits
Next, we need to change the limits of integration. When \( x = 0 \):
\[ t = 1 + \sin(0) = 1. \]
When \( x = \frac{\pi}{2} \):
\[ t = 1 + \sin\left(\frac{\pi}{2}\right) = 1 + 1 = 2. \]
Thus, the new limits for \( t \) are from 1 to 2.

### Step 3: Express \(\cos x\) in terms of \(t\)
From our substitution \( t = 1 + \sin x \), we can express \(\sin x\) as:
\[ \sin x = t - 1. \]
Using the identity \(\cos^2 x + \sin^2 x = 1\), we have:
\[ \cos^2 x = 1 - \sin^2 x = 1 - (t - 1)^2 = 1 - (t^2 - 2t + 1) = 2t - t^2. \]
Thus,
\[ \cos x = \sqrt{2t - t^2}. \]

### Step 4: Substitute in the Integral
Now, substituting \( \cos x \) and \( dx \) into the integral, we get:
\[ I = \int_{1}^{2} \frac{\sqrt{2t - t^2}}{t^3} \cdot \frac{dt}{\sqrt{2t - t^2}} = \int_{1}^{2} \frac{1}{t^3} \, dt. \]

### Step 5: Integrate
Now we can integrate:
\[ I = \int_{1}^{2} t^{-3} \, dt = \left[ -\frac{1}{2t^2} \right]_{1}^{2}. \]
Calculating the limits:
\[ I = -\frac{1}{2(2^2)} - \left(-\frac{1}{2(1^2)}\right) = -\frac{1}{8} + \frac{1}{2} = -\frac{1}{8} + \frac{4}{8} = \frac{3}{8}. \]

### Final Answer
Thus, the value of the integral is:
\[ I = \frac{3}{8}. \]