If `2int_0^1 tan^(-1)x dx= int_0^1 cot^(-1)(1-x+x^2) dx` then `int_0^1 tan^(-1)(1-x+x^2) dx=`

Answer

To solve the problem, we start with the given equation:

\[ 2 \int_0^1 \tan^{-1}(x) \, dx = \int_0^1 \cot^{-1}(1 - x + x^2) \, dx \]

We need to find the value of:

\[ \int_0^1 \tan^{-1}(1 - x + x^2) \, dx \]

Let's denote:

\[ I = \int_0^1 \tan^{-1}(1 - x + x^2) \, dx \]

Using the identity:

\[ \tan^{-1}(a) + \cot^{-1}(a) = \frac{\pi}{2} \]

we can express \( I \) as:

\[ I = \int_0^1 \left( \frac{\pi}{2} - \cot^{-1}(1 - x + x^2) \right) \, dx \]

This gives us:

\[ I = \frac{\pi}{2} \int_0^1 dx - \int_0^1 \cot^{-1}(1 - x + x^2) \, dx \]

Calculating the first integral:

\[ \int_0^1 dx = 1 \]

Thus, we have:

\[ I = \frac{\pi}{2} - \int_0^1 \cot^{-1}(1 - x + x^2) \, dx \]

From the problem statement, we know:

\[ \int_0^1 \cot^{-1}(1 - x + x^2) \, dx = 2 \int_0^1 \tan^{-1}(x) \, dx \]

Let’s denote:

\[ J = \int_0^1 \tan^{-1}(x) \, dx \]

So we can rewrite \( I \):

\[ I = \frac{\pi}{2} - 2J \]

Next, we need to calculate \( J \). We can use integration by parts for \( J \):

Let \( u = \tan^{-1}(x) \) and \( dv = dx \). Then:

\[ du = \frac{1}{1 + x^2} \, dx \quad \text{and} \quad v = x \]

Using integration by parts:

\[ J = \left[ x \tan^{-1}(x) \right]_0^1 - \int_0^1 \frac{x}{1 + x^2} \, dx \]

Calculating the boundary term:

\[ \left[ x \tan^{-1}(x) \right]_0^1 = 1 \cdot \tan^{-1}(1) - 0 \cdot \tan^{-1}(0) = \frac{\pi}{4} \]

Now we calculate the second integral:

\[ \int_0^1 \frac{x}{1 + x^2} \, dx \]

Using substitution \( t = 1 + x^2 \), \( dt = 2x \, dx \):

When \( x = 0 \), \( t = 1 \) and when \( x = 1 \), \( t = 2 \):

\[ \int_0^1 \frac{x}{1 + x^2} \, dx = \frac{1}{2} \int_1^2 \frac{1}{t} \, dt = \frac{1}{2} \left[ \ln(t) \right]_1^2 = \frac{1}{2} (\ln(2) - \ln(1)) = \frac{1}{2} \ln(2) \]

Putting this back into \( J \):

\[ J = \frac{\pi}{4} - \frac{1}{2} \ln(2) \]

Now substituting \( J \) back into the equation for \( I \):

\[ I = \frac{\pi}{2} - 2 \left( \frac{\pi}{4} - \frac{1}{2} \ln(2) \right) \]

This simplifies to:

\[ I = \frac{\pi}{2} - \frac{\pi}{2} + \ln(2) = \ln(2) \]

Thus, the final result is:

\[ \int_0^1 \tan^{-1}(1 - x + x^2) \, dx = \ln(2) \]