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

Options

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

Answer

To solve the integral \( \int \frac{dx}{\sqrt{1 - \tan^2 x}} \) and find the value of \( \lambda \) such that

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

we can follow these steps:

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

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

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

\[ 1 - \tan^2 x = 1 - \frac{\sin^2 x}{\cos^2 x} = \frac{\cos^2 x - \sin^2 x}{\cos^2 x} = \frac{\cos^2 x - \sin^2 x}{\cos^2 x}. \]

Thus, we have:

\[ \sqrt{1 - \tan^2 x} = \sqrt{\frac{\cos^2 x - \sin^2 x}{\cos^2 x}} = \frac{\sqrt{\cos^2 x - \sin^2 x}}{\cos x}. \]

### Step 2: Substitute into the Integral
Substituting this back into the integral gives:

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

### Step 3: Use a Trigonometric Identity
We know that \( \cos^2 x + \sin^2 x = 1 \). Therefore, we can express \( \cos^2 x - \sin^2 x \) as:

\[ \cos^2 x - \sin^2 x = \cos 2x. \]

So we can rewrite our integral as:

\[ \int \frac{\cos x \, dx}{\sqrt{\cos 2x}}. \]

### Step 4: Use Substitution
Let \( t = \sin x \). Then, \( dt = \cos x \, dx \). The integral becomes:

\[ \int \frac{dt}{\sqrt{1 - t^2}}. \]

### Step 5: Solve the Integral
The integral \( \int \frac{dt}{\sqrt{1 - t^2}} \) is known to be:

\[ \sin^{-1}(t) + C. \]

### Step 6: Back Substitute
Substituting back \( t = \sin x \):

\[ \sin^{-1}(\sin x) + C = x + C. \]

### Step 7: Compare with Given Expression
Now, we need to compare this result with the given expression:

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

From the expression, we can see that:

\[ \sin^{-1}(\lambda \sin x) = \lambda \sin^{-1}(\sin x). \]

### Step 8: Find \( \lambda \)
To match the coefficients, we need:

\[ \frac{1}{\lambda} = \frac{1}{\sqrt{2}} \implies \lambda = \sqrt{2}. \]

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

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