`int_(0)^(pi//2)sin^(3)x cos^(4)x dx=`

Options

A. `(2)/(35)`
B. `(35)/(2)`
C. `(35pi)/(2)`
D. `(2pi)/(35)`

Answer

To solve the integral \( I = \int_{0}^{\frac{\pi}{2}} \sin^3 x \cos^4 x \, dx \), we can follow these steps:

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

### Step 2: Substitute \(\sin^2 x\)
Using the identity \(\sin^2 x = 1 - \cos^2 x\), we can rewrite the integral:
\[ I = \int_{0}^{\frac{\pi}{2}} \sin x \cdot (1 - \cos^2 x) \cdot \cos^4 x \, dx \]
This expands to:
\[ I = \int_{0}^{\frac{\pi}{2}} \sin x \cdot \cos^4 x \, dx - \int_{0}^{\frac{\pi}{2}} \sin x \cdot \cos^6 x \, dx \]

### Step 3: Change of Variables
For both integrals, we can use the substitution \( t = \cos x \). Then, \( dt = -\sin x \, dx \). The limits change as follows:
- When \( x = 0 \), \( t = 1 \)
- When \( x = \frac{\pi}{2} \), \( t = 0 \)

Thus, we have:
\[ I = -\int_{1}^{0} t^4 (-dt) - \int_{1}^{0} t^6 (-dt) \]
This simplifies to:
\[ I = \int_{0}^{1} t^4 \, dt - \int_{0}^{1} t^6 \, dt \]

### Step 4: Evaluate the Integrals
Now we can evaluate each integral:
1. For \(\int_{0}^{1} t^4 \, dt\):
\[ \int t^4 \, dt = \frac{t^5}{5} \Big|_{0}^{1} = \frac{1^5}{5} - \frac{0^5}{5} = \frac{1}{5} \]

2. For \(\int_{0}^{1} t^6 \, dt\):
\[ \int t^6 \, dt = \frac{t^7}{7} \Big|_{0}^{1} = \frac{1^7}{7} - \frac{0^7}{7} = \frac{1}{7} \]

### Step 5: Combine the Results
Now substituting back, we get:
\[ I = \frac{1}{5} - \frac{1}{7} \]
To combine these fractions, we find a common denominator:
\[ I = \frac{7}{35} - \frac{5}{35} = \frac{2}{35} \]

### Final Answer
Thus, the value of the integral is:
\[ \int_{0}^{\frac{\pi}{2}} \sin^3 x \cos^4 x \, dx = \frac{2}{35} \]