`int_(0)^(pi//2) sqrt(cos x) sin ^(3) x" "dx=`

Options

A. `4/21`
B. `10/21`
C. `8/21`
D. `20/21`

Answer

To solve the integral \( I = \int_{0}^{\frac{\pi}{2}} \sqrt{\cos x} \sin^3 x \, dx \), we can follow these steps:

### Step 1: Rewrite the Integral
We can express \( \sin^3 x \) as \( \sin^2 x \cdot \sin x \). Thus, we rewrite the integral as:
\[ I = \int_{0}^{\frac{\pi}{2}} \sqrt{\cos x} \sin^2 x \sin x \, dx \]

### Step 2: Use the Identity for \( \sin^2 x \)
Using the identity \( \sin^2 x = 1 - \cos^2 x \), we can substitute:
\[ I = \int_{0}^{\frac{\pi}{2}} \sqrt{\cos x} (1 - \cos^2 x) \sin x \, dx \]
This simplifies to:
\[ I = \int_{0}^{\frac{\pi}{2}} \sqrt{\cos x} \sin x \, dx - \int_{0}^{\frac{\pi}{2}} \sqrt{\cos x} \cos^2 x \sin x \, dx \]

### Step 3: Substitution
Let \( t = \cos x \). Then, \( dt = -\sin x \, dx \). The limits change as follows:
- When \( x = 0 \), \( t = \cos(0) = 1 \)
- When \( x = \frac{\pi}{2} \), \( t = \cos\left(\frac{\pi}{2}\right) = 0 \)

Thus, we can rewrite the integral:
\[ I = -\int_{1}^{0} \sqrt{t} (1 - t^2) \, dt \]
Reversing the limits gives:
\[ I = \int_{0}^{1} \sqrt{t} (1 - t^2) \, dt \]

### Step 4: Expand the Integral
Now we expand the integrand:
\[ I = \int_{0}^{1} \left( \sqrt{t} - t^{5/2} \right) \, dt \]

### Step 5: Integrate
Now we can integrate term by term:
\[ I = \int_{0}^{1} t^{1/2} \, dt - \int_{0}^{1} t^{5/2} \, dt \]
Calculating these integrals:
\[ \int t^{1/2} \, dt = \frac{t^{3/2}}{3/2} = \frac{2}{3} t^{3/2} \quad \text{from } 0 \text{ to } 1 \]
\[ \int t^{5/2} \, dt = \frac{t^{7/2}}{7/2} = \frac{2}{7} t^{7/2} \quad \text{from } 0 \text{ to } 1 \]
Thus:
\[ I = \left[ \frac{2}{3} \cdot 1^{3/2} \right] - \left[ \frac{2}{7} \cdot 1^{7/2} \right] = \frac{2}{3} - \frac{2}{7} \]

### Step 6: Common Denominator
Finding a common denominator (which is 21):
\[ I = \frac{14}{21} - \frac{6}{21} = \frac{8}{21} \]

### Final Answer
Thus, the value of the integral is:
\[ I = \frac{8}{21} \]

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