The integral `I=int(sin^(3)thetacos theta)/((1+sin^(2)theta)^(2))d theta` simplifies to (where, c is the constant of integration)

Options

A. `(1)/(2)ln(sin theta)+(1)/(1+sin^(2)theta)+c`
B. `(1)/(2)[ln(1+sin^(2)theta)+(1)/(1+sin^(2)theta)]+c`
C. `ln(sin theta)+(1)/(1+sin^(2)theta)+c`
D. `ln(sin^(2)theta+1)+(1)/(sin^(2)theta+2)+c`

Answer

To solve the integral

\[ I = \int \frac{\sin^3 \theta \cos \theta}{(1 + \sin^2 \theta)^2} \, d\theta, \]

we can follow these steps:

### Step 1: Rewrite the Integral
We can express \(\sin^3 \theta\) in terms of \(\sin^2 \theta\):
\[ \sin^3 \theta = \sin^2 \theta \cdot \sin \theta = (1 - \cos^2 \theta) \sin \theta. \]
Thus, we rewrite the integral:
\[ I = \int \frac{(1 - \cos^2 \theta) \sin \theta \cos \theta}{(1 + \sin^2 \theta)^2} \, d\theta. \]

### Step 2: Substitution
Let \(T = 1 + \sin^2 \theta\). Then, differentiating both sides gives:
\[ dT = 2 \sin \theta \cos \theta \, d\theta \quad \Rightarrow \quad \sin \theta \cos \theta \, d\theta = \frac{dT}{2}. \]
Now, we need to express \(\sin^2 \theta\) in terms of \(T\):
\[ \sin^2 \theta = T - 1. \]

### Step 3: Substitute in the Integral
Substituting \(T\) and \(dT\) into the integral, we have:
\[ I = \int \frac{(T - 1) \cdot \frac{dT}{2}}{T^2}. \]
This simplifies to:
\[ I = \frac{1}{2} \int \frac{T - 1}{T^2} \, dT. \]

### Step 4: Break Down the Integral
We can break this integral into two parts:
\[ I = \frac{1}{2} \left( \int \frac{1}{T} \, dT - \int \frac{1}{T^2} \, dT \right). \]

### Step 5: Integrate Each Part
Now we can integrate each part:
1. \(\int \frac{1}{T} \, dT = \ln |T|\)
2. \(\int \frac{1}{T^2} \, dT = -\frac{1}{T}\)

Thus, we have:
\[ I = \frac{1}{2} \left( \ln |T| + \frac{1}{T} \right) + C. \]

### Step 6: Substitute Back for \(T\)
Now substitute back \(T = 1 + \sin^2 \theta\):
\[ I = \frac{1}{2} \left( \ln(1 + \sin^2 \theta) + \frac{1}{1 + \sin^2 \theta} \right) + C. \]

### Final Result
Therefore, the integral simplifies to:
\[ I = \frac{1}{2} \ln(1 + \sin^2 \theta) + \frac{1}{2(1 + \sin^2 \theta)} + C. \]