Evaluate `inttheta^(-(1)/(2)) sin theta^((1)/(2)) d theta.`

Answer

To evaluate the integral \( \int \theta^{-\frac{1}{2}} \sin(\theta^{\frac{1}{2}}) \, d\theta \), we will use substitution. Here’s a step-by-step solution:

### Step 1: Choose a substitution
Let \( t = \theta^{\frac{1}{2}} \). Then, we have:
\[ \theta = t^2 \]
Now, differentiate both sides with respect to \( \theta \):
\[ d\theta = 2t \, dt \]

### Step 2: Rewrite the integral in terms of \( t \)
Substituting \( \theta = t^2 \) and \( d\theta = 2t \, dt \) into the integral, we get:
\[ \int \theta^{-\frac{1}{2}} \sin(\theta^{\frac{1}{2}}) \, d\theta = \int (t^2)^{-\frac{1}{2}} \sin(t) \cdot (2t \, dt) \]
This simplifies to:
\[ = \int t^{-1} \sin(t) \cdot (2t \, dt) = 2 \int \sin(t) \, dt \]

### Step 3: Integrate
Now we can integrate:
\[ 2 \int \sin(t) \, dt = -2 \cos(t) + C \]

### Step 4: Substitute back to \( \theta \)
Recall that \( t = \theta^{\frac{1}{2}} \), so substituting back gives:
\[ -2 \cos(\theta^{\frac{1}{2}}) + C \]

### Final Answer
Thus, the evaluated integral is:
\[ \int \theta^{-\frac{1}{2}} \sin(\theta^{\frac{1}{2}}) \, d\theta = -2 \cos(\theta^{\frac{1}{2}}) + C \]

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