The value of `intsin^(3)x sqrt(cosx)dx` is equal to (where, c is the constant of integration)

Options

A. `(2)/(3)(cosx)^((3)/(2))+(2)/(7)(cosx)^((7)/(2))+c`
B. `-(2)/(3)(cosx)^((3)/(2))+(2)/(7)(cosx)^((7)/(2))+c`
C. `(-(2)/(3))(cosx)^(3)+(2)/(5)(cosx^((5)/(2)))+c`
D. `(3)/(2)(cosx)^((3)/(2))+(5)/(2)(cosx)^((7)/(2))+c`

Answer

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

### Step 1: Rewrite the Integral
We start by rewriting \( \sin^3 x \) as \( \sin^2 x \cdot \sin x \):
\[ I = \int \sin^2 x \cdot \sin x \cdot \sqrt{\cos x} \, dx \]

### Step 2: Use the Pythagorean Identity
Using the identity \( \sin^2 x = 1 - \cos^2 x \), we can substitute:
\[ I = \int (1 - \cos^2 x) \sin x \cdot \sqrt{\cos x} \, dx \]

### Step 3: Substitute for \( \cos x \)
Let \( \cos x = t \). Then, differentiating gives:
\[ -\sin x \, dx = dt \quad \Rightarrow \quad \sin x \, dx = -dt \]
Now, substituting into the integral:
\[ I = \int (1 - t^2) \sqrt{t} (-dt) = -\int (1 - t^2) t^{1/2} \, dt \]

### Step 4: Simplify the Integral
Distributing \( t^{1/2} \):
\[ I = -\int (t^{1/2} - t^{5/2}) \, dt \]

### Step 5: Integrate Each Term
Now, we can integrate each term:
\[ I = -\left( \frac{t^{3/2}}{3/2} - \frac{t^{7/2}}{7/2} \right) + C \]
This simplifies to:
\[ I = -\left( \frac{2}{3} t^{3/2} - \frac{2}{7} t^{7/2} \right) + C \]

### Step 6: Substitute Back for \( t \)
Substituting back \( t = \cos x \):
\[ I = -\left( \frac{2}{3} \cos^{3/2} x - \frac{2}{7} \cos^{7/2} x \right) + C \]
This gives us:
\[ I = \frac{2}{3} \cos^{3/2} x - \frac{2}{7} \cos^{7/2} x + C \]

### Final Answer
Thus, the value of the integral is:
\[ I = -\frac{2}{3} \cos^{3/2} x + \frac{2}{7} \cos^{7/2} x + C \]