`int (dx)/(sqrt(1-tan ^(2) x))=(1)/(lamda)sin ^(-1) (lamda sin x)+C, then lamda=`

Options

A. `sqrt2`
B. `sqrt3`
C. 2
D. `sqrt5`

Answer

To solve the integral

\[ \int \frac{dx}{\sqrt{1 - \tan^2 x}} = \frac{1}{\lambda} \sin^{-1}(\lambda \sin x) + C \]

we will follow these steps:

### Step 1: Simplify the Integral
We start with the integral:

\[ \int \frac{dx}{\sqrt{1 - \tan^2 x}} \]

Using the identity \(1 - \tan^2 x = \frac{\cos^2 x - \sin^2 x}{\cos^2 x}\), we can rewrite the integral as:

\[ \int \frac{dx}{\sqrt{\frac{\cos^2 x - \sin^2 x}{\cos^2 x}}} = \int \frac{\cos x \, dx}{\sqrt{\cos^2 x - \sin^2 x}} \]

### Step 2: Use a Trigonometric Substitution
Next, we can simplify the expression under the square root:

\[ \sqrt{\cos^2 x - \sin^2 x} = \sqrt{\cos^2 x(1 - \tan^2 x)} = \cos x \sqrt{1 - \tan^2 x} \]

Thus, we have:

\[ \int \frac{\cos x \, dx}{\cos x \sqrt{1 - \tan^2 x}} = \int \frac{dx}{\sqrt{1 - \tan^2 x}} \]

### Step 3: Change of Variables
Now, we can use the substitution \(t = \sqrt{2} \sin x\). Then, differentiating gives:

\[ dt = \sqrt{2} \cos x \, dx \quad \Rightarrow \quad dx = \frac{dt}{\sqrt{2} \cos x} \]

Substituting this into the integral gives:

\[ \int \frac{1}{\sqrt{1 - \left(\frac{t}{\sqrt{2}}\right)^2}} \cdot \frac{dt}{\sqrt{2} \cos x} \]

### Step 4: Evaluate the Integral
The integral simplifies to:

\[ \frac{1}{\sqrt{2}} \int \frac{dt}{\sqrt{1 - t^2}} = \frac{1}{\sqrt{2}} \sin^{-1}(t) + C \]

Substituting back \(t = \sqrt{2} \sin x\):

\[ \frac{1}{\sqrt{2}} \sin^{-1}(\sqrt{2} \sin x) + C \]

### Step 5: Compare with Given Expression
Now, we compare this with the expression given in the problem:

\[ \frac{1}{\lambda} \sin^{-1}(\lambda \sin x) + C \]

From this comparison, we can see that:

\[ \frac{1}{\lambda} = \frac{1}{\sqrt{2}} \quad \text{and} \quad \lambda = \sqrt{2} \]

### Conclusion
Thus, the value of \(\lambda\) is:

\[ \lambda = \sqrt{2} \]