`int_(0)^(pi//2) sin^(4)xcos^(3)dx` is equal to :

Options

A. `6/35`
B. `2/21`
C. `2/15`
D. `2/35`

Answer

To solve the integral \( I = \int_{0}^{\frac{\pi}{2}} \sin^4 x \cos^3 x \, dx \), we will follow a systematic approach.

### Step 1: Rewrite the integral
We can express \( \cos^3 x \) as \( \cos^2 x \cdot \cos x \). Thus, we can rewrite the integral as:
\[ I = \int_{0}^{\frac{\pi}{2}} \sin^4 x \cos^2 x \cos x \, dx = \int_{0}^{\frac{\pi}{2}} \sin^4 x (1 - \sin^2 x) \cos x \, dx \]

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

Now, the integral becomes:
\[ I = \int_{0}^{1} t^4 (1 - t^2) \, dt \]

### Step 3: Expand the integrand
Now we can expand the integrand:
\[ I = \int_{0}^{1} (t^4 - t^6) \, dt \]

### Step 4: Integrate term by term
Now we can integrate each term separately:
\[ I = \int_{0}^{1} t^4 \, dt - \int_{0}^{1} t^6 \, dt \]

Using the formula for integration \( \int t^n \, dt = \frac{t^{n+1}}{n+1} \):
\[ \int_{0}^{1} t^4 \, dt = \frac{1^5}{5} - \frac{0^5}{5} = \frac{1}{5} \]
\[ \int_{0}^{1} t^6 \, dt = \frac{1^7}{7} - \frac{0^7}{7} = \frac{1}{7} \]

### Step 5: Combine the results
Now substituting back into our expression for \( I \):
\[ I = \frac{1}{5} - \frac{1}{7} \]

### Step 6: Find a common denominator
The common denominator for 5 and 7 is 35:
\[ I = \frac{7}{35} - \frac{5}{35} = \frac{2}{35} \]

### Final Answer
Thus, the value of the integral is:
\[ I = \frac{2}{35} \]