If `y=int1/(1+x^2)^(3/2)dx` and `y=0` when `x=0` , then value of y when `x=1` is

Options

A. `(1)/(sqrt(2))`
B. `sqrt(2)`
C. `2sqrt(2)`
D. none of these

Explanation

To solve the problem step by step, we start with the given integral:

### Step 1: Set up the integral
We have:
\[ y = \int \frac{1}{(1 + x^2)^{3/2}} \, dx \]

### Step 2: Use a trigonometric substitution
Let \( x = \tan(\theta) \). Then, the differential \( dx \) becomes:
\[ dx = \sec^2(\theta) \, d\theta \]
Also, we know that:
\[ 1 + x^2 = 1 + \tan^2(\theta) = \sec^2(\theta) \]
Thus, we can rewrite the integral as:
\[ y = \int \frac{1}{(\sec^2(\theta))^{3/2}} \sec^2(\theta) \, d\theta \]

### Step 3: Simplify the integral
We simplify the expression:
\[ (\sec^2(\theta))^{3/2} = \sec^3(\theta) \]
So the integral becomes:
\[ y = \int \frac{\sec^2(\theta)}{\sec^3(\theta)} \, d\theta = \int \cos(\theta) \, d\theta \]

### Step 4: Integrate
The integral of \( \cos(\theta) \) is:
\[ y = \sin(\theta) + C \]

### Step 5: Back-substitute for \( \theta \)
Since \( \theta = \tan^{-1}(x) \), we have:
\[ y = \sin(\tan^{-1}(x)) + C \]

### Step 6: Find the value of \( \sin(\tan^{-1}(x)) \)
Using the identity for sine:
\[ \sin(\tan^{-1}(x)) = \frac{x}{\sqrt{1 + x^2}} \]
Thus, we can write:
\[ y = \frac{x}{\sqrt{1 + x^2}} + C \]

### Step 7: Apply the initial condition
We know that \( y = 0 \) when \( x = 0 \):
\[ 0 = \frac{0}{\sqrt{1 + 0^2}} + C \implies C = 0 \]
So we have:
\[ y = \frac{x}{\sqrt{1 + x^2}} \]

### Step 8: Find \( y \) when \( x = 1 \)
Now, we substitute \( x = 1 \):
\[ y = \frac{1}{\sqrt{1 + 1^2}} = \frac{1}{\sqrt{2}} = \frac{1}{\sqrt{2}} \]

### Final Answer
Thus, the value of \( y \) when \( x = 1 \) is:
\[ \frac{1}{\sqrt{2}} \]

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