Evaluate :
`int sin^(-1)((6x)/(1+9x^(2)))dx`

Answer

To evaluate the integral \( I = \int \sin^{-1}\left(\frac{6x}{1 + 9x^2}\right) dx \), we will follow a systematic approach.

### Step 1: Simplify the Argument of the Inverse Sine Function

We start by rewriting the argument of the inverse sine function:
\[ \frac{6x}{1 + 9x^2} = \frac{2 \cdot 3x}{1 + (3x)^2} \]
This suggests a trigonometric substitution. Let \( 3x = \tan \theta \), then \( x = \frac{1}{3} \tan \theta \).

### Step 2: Change of Variables

Differentiating \( x = \frac{1}{3} \tan \theta \) gives:
\[ dx = \frac{1}{3} \sec^2 \theta \, d\theta \]
Now substitute \( 3x = \tan \theta \) into the integral:
\[ I = \int \sin^{-1}\left(\frac{2 \tan \theta}{1 + \tan^2 \theta}\right) \cdot \frac{1}{3} \sec^2 \theta \, d\theta \]

### Step 3: Simplifying the Inverse Sine Expression

Using the identity \( \frac{2 \tan \theta}{1 + \tan^2 \theta} = \sin(2\theta) \), we can simplify:
\[ \sin^{-1}\left(\frac{2 \tan \theta}{1 + \tan^2 \theta}\right) = \sin^{-1}(\sin(2\theta)) = 2\theta \]
Thus, we can rewrite the integral:
\[ I = \int 2\theta \cdot \frac{1}{3} \sec^2 \theta \, d\theta = \frac{2}{3} \int \theta \sec^2 \theta \, d\theta \]

### Step 4: Integration by Parts

Now we apply integration by parts. Let:
- \( u = \theta \) (thus \( du = d\theta \))
- \( dv = \sec^2 \theta \, d\theta \) (thus \( v = \tan \theta \))

Using the integration by parts formula \( \int u \, dv = uv - \int v \, du \):
\[ \int \theta \sec^2 \theta \, d\theta = \theta \tan \theta - \int \tan \theta \, d\theta \]
The integral of \( \tan \theta \) is \( -\log|\cos \theta| \), so:
\[ \int \theta \sec^2 \theta \, d\theta = \theta \tan \theta + \log|\cos \theta| \]

### Step 5: Substitute Back

Substituting back into our expression for \( I \):
\[ I = \frac{2}{3} \left( \theta \tan \theta + \log|\cos \theta| \right) + C \]
Now substituting \( \theta = \tan^{-1}(3x) \) and \( \tan \theta = 3x \):
\[ I = \frac{2}{3} \left( \tan^{-1}(3x) \cdot 3x + \log|\cos(\tan^{-1}(3x))| \right) + C \]

### Step 6: Simplifying the Logarithmic Term

Using the identity \( \cos(\tan^{-1}(3x)) = \frac{1}{\sqrt{1 + (3x)^2}} = \frac{1}{\sqrt{1 + 9x^2}} \):
\[ \log|\cos(\tan^{-1}(3x))| = \log\left(\frac{1}{\sqrt{1 + 9x^2}}\right) = -\frac{1}{2} \log(1 + 9x^2) \]

### Final Expression

Thus, the final result for the integral is:
\[ I = \frac{2}{3} \left( 3x \tan^{-1}(3x) - \frac{1}{2} \log(1 + 9x^2) \right) + C \]
This simplifies to:
\[ I = 2x \tan^{-1}(3x) - \frac{1}{3} \log(1 + 9x^2) + C \]