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

Options

A. 35
B. `(1)/(35)`
C. `(2)/(35)`
D. `(1)/(7)`

Answer

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

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

### Step 2: Substitution
Now, we will use the substitution \( t = \sin x \). Then, we differentiate:
\[ dt = \cos x \, dx \quad \Rightarrow \quad dx = \frac{dt}{\cos x} \]
We also need to change the limits of integration. When \( x = 0 \), \( t = \sin(0) = 0 \) and when \( x = \frac{\pi}{2} \), \( t = \sin\left(\frac{\pi}{2}\right) = 1 \). Therefore, the limits change from \( 0 \) to \( 1 \).

### Step 3: Substitute in the Integral
Substituting \( t \) into the integral gives:
\[ \int_{0}^{1} t^4 (1 - t^2) \, dt \]
This simplifies to:
\[ \int_{0}^{1} (t^4 - t^6) \, dt \]

### Step 4: Integrate
Now we can integrate term by term:
\[ \int_{0}^{1} t^4 \, dt - \int_{0}^{1} t^6 \, dt \]
Calculating these integrals:
\[ \int t^4 \, dt = \frac{t^5}{5} \quad \text{and} \quad \int t^6 \, dt = \frac{t^7}{7} \]
Thus, we have:
\[ \left[ \frac{t^5}{5} \right]_{0}^{1} - \left[ \frac{t^7}{7} \right]_{0}^{1} \]
Evaluating at the limits:
\[ \left( \frac{1^5}{5} - 0 \right) - \left( \frac{1^7}{7} - 0 \right) = \frac{1}{5} - \frac{1}{7} \]

### Step 5: Find a Common Denominator
To combine these fractions, we find a common denominator, which is \( 35 \):
\[ \frac{1}{5} = \frac{7}{35} \quad \text{and} \quad \frac{1}{7} = \frac{5}{35} \]
Thus:
\[ \frac{1}{5} - \frac{1}{7} = \frac{7}{35} - \frac{5}{35} = \frac{2}{35} \]

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

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