`int dt/sqrt[3t-2t^2]`

Explanation

To solve the integral \( I = \int \frac{dt}{\sqrt{3t - 2t^2}} \), we can follow these steps:

### Step 1: Factor out constants
We start by rewriting the integral:
\[ I = \int \frac{dt}{\sqrt{3t - 2t^2}} = \int \frac{dt}{\sqrt{-2t^2 + 3t}} \]
Next, we factor out \(-2\) from the expression inside the square root:
\[ I = \int \frac{dt}{\sqrt{-2(t^2 - \frac{3}{2}t)}} = \int \frac{dt}{\sqrt{-2\left(t^2 - \frac{3}{2}t\right)}} \]

### Step 2: Complete the square
Now, we complete the square for the quadratic expression inside the square root:
\[ t^2 - \frac{3}{2}t = \left(t - \frac{3}{4}\right)^2 - \left(\frac{3}{4}\right)^2 \]
Thus, we have:
\[ I = \int \frac{dt}{\sqrt{-2\left(\left(t - \frac{3}{4}\right)^2 - \left(\frac{3}{4}\right)^2\right)}} \]

### Step 3: Simplify the integral
Substituting this back into the integral gives:
\[ I = \int \frac{dt}{\sqrt{-2\left(t - \frac{3}{4}\right)^2 + \frac{9}{8}}} \]
This can be rewritten as:
\[ I = \int \frac{dt}{\sqrt{\frac{9}{8} - 2\left(t - \frac{3}{4}\right)^2}} \]

### Step 4: Use trigonometric substitution
We can use the substitution \( u = t - \frac{3}{4} \), which gives \( dt = du \):
\[ I = \int \frac{du}{\sqrt{\frac{9}{8} - 2u^2}} \]
Now, we can factor out constants:
\[ I = \int \frac{du}{\sqrt{\frac{9}{8}} \sqrt{1 - \frac{16}{9}u^2}} = \frac{1}{\sqrt{\frac{9}{8}}} \int \frac{du}{\sqrt{1 - \left(\frac{4u}{3}\right)^2}} \]

### Step 5: Evaluate the integral
Using the integral formula \( \int \frac{dx}{\sqrt{1 - x^2}} = \sin^{-1}(x) + C \):
\[ I = \frac{1}{\sqrt{\frac{9}{8}}} \cdot \sin^{-1}\left(\frac{4u}{3}\right) + C \]
Substituting back \( u = t - \frac{3}{4} \):
\[ I = \frac{1}{\sqrt{\frac{9}{8}}} \cdot \sin^{-1}\left(\frac{4(t - \frac{3}{4})}{3}\right) + C \]

### Step 6: Final simplification
Simplifying gives:
\[ I = \frac{2\sqrt{2}}{3} \sin^{-1}\left(\frac{4t - 3}{3}\right) + C \]

### Final Answer:
\[ I = \frac{2\sqrt{2}}{3} \sin^{-1}\left(\frac{4t - 3}{3}\right) + C \]