Find the error in steps to evaluate the following integral `int_(0)^(pi)(dx)/(1+2 sin ^(2) x )=int _(0)^(pi)(sec^(2)xdx)/(sec^(2)x+2 tan^(2)x)=int_(0)^(pi) (sec^(2)xdx)/(1+3 tan^(2)x)`
`=(1)/(sqrt3)[tan^(-1)(sqrt3 tan x)]_(0)^(pi)=0`

Answer

To evaluate the integral

\[ I = \int_{0}^{\pi} \frac{dx}{1 + 2 \sin^2 x} \]

we will go through the steps carefully and identify any errors in the provided solution.

### Step 1: Rewrite the Integral

We start with the integral:

\[ I = \int_{0}^{\pi} \frac{dx}{1 + 2 \sin^2 x} \]

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

\[ I = \int_{0}^{\pi} \frac{dx}{1 + 2 \cdot \frac{1 - \cos(2x)}{2}} = \int_{0}^{\pi} \frac{dx}{2 - \cos(2x)} \]

### Step 2: Use a Trigonometric Substitution

Next, we can use the substitution \(t = \tan x\). Then, \(dx = \frac{dt}{1 + t^2}\) and the limits change from \(x = 0\) to \(x = \pi\) which corresponds to \(t = 0\) to \(t = \infty\).

Substituting into the integral gives:

\[ I = \int_{0}^{\infty} \frac{dt}{(2 - \cos(2 \tan^{-1} t))(1 + t^2)} \]

### Step 3: Simplify the Integral

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

\[ I = \int_{0}^{\infty} \frac{dt}{2 - \frac{1 - t^2}{1 + t^2}} \cdot \frac{1}{1 + t^2} \]

This simplifies to:

\[ I = \int_{0}^{\infty} \frac{(1 + t^2) dt}{2(1 + t^2) - (1 - t^2)} = \int_{0}^{\infty} \frac{(1 + t^2) dt}{3 + t^2} \]

### Step 4: Evaluate the Integral

Now we can evaluate the integral:

\[ I = \int_{0}^{\infty} \frac{(1 + t^2) dt}{3 + t^2} = \int_{0}^{\infty} \frac{dt}{3 + t^2} + \int_{0}^{\infty} \frac{t^2 dt}{3 + t^2} \]

The first integral can be evaluated using the formula \(\int \frac{dx}{a^2 + x^2} = \frac{1}{a} \tan^{-1} \left( \frac{x}{a} \right)\):

\[ \int_{0}^{\infty} \frac{dt}{3 + t^2} = \frac{1}{\sqrt{3}} \cdot \frac{\pi}{2} = \frac{\pi}{2\sqrt{3}} \]

The second integral can be evaluated by using integration by parts or recognizing it as a standard integral:

\[ \int_{0}^{\infty} \frac{t^2 dt}{3 + t^2} = \frac{1}{2} \int_{0}^{\infty} \frac{d(3 + t^2)}{3 + t^2} = \frac{1}{2} \cdot \frac{\pi}{\sqrt{3}} = \frac{\pi}{6} \]

### Final Step: Combine Results

Thus, we have:

\[ I = \frac{\pi}{2\sqrt{3}} + \frac{\pi}{6} \]

Combining these gives us:

\[ I = \frac{3\pi + \pi}{6\sqrt{3}} = \frac{4\pi}{6\sqrt{3}} = \frac{2\pi}{3\sqrt{3}} \]

### Conclusion

The error in the original steps was in the evaluation of the limits and the final integration. The final result is not zero, but rather:

\[ I = \frac{2\pi}{3\sqrt{3}} \]