`int(sin^(-1)sqrtx-cos^(-1)sqrtx)/(sin^(-1)sqrtx+cos^(-1)sqrtx)dx=`

Options

A. `2/pi[sqrt((x-x^2))+(1-2x)sin^(-1)sqrtx]+x`
B. `2/pi[sqrt((x-x^2))-(1-2x)sin^(-1)sqrtx]-x`
C. `2/pi[sqrt((x-x^2))+(1-2x)sin^(-1)sqrtx]-x`
D. None

Answer

To solve the integral

\[ I = \int \frac{\sin^{-1}(\sqrt{x}) - \cos^{-1}(\sqrt{x})}{\sin^{-1}(\sqrt{x}) + \cos^{-1}(\sqrt{x})} \, dx \]

we start by recognizing a key identity involving inverse trigonometric functions. We know that:

\[ \sin^{-1}(y) + \cos^{-1}(y) = \frac{\pi}{2} \]

for any \(y\). In our case, \(y = \sqrt{x}\). Therefore, we can rewrite the denominator:

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

Substituting this into our integral, we have:

\[ I = \int \frac{\sin^{-1}(\sqrt{x}) - \cos^{-1}(\sqrt{x})}{\frac{\pi}{2}} \, dx = \frac{2}{\pi} \int \left( \sin^{-1}(\sqrt{x}) - \cos^{-1}(\sqrt{x}) \right) \, dx \]

Next, we will break this integral into two parts:

\[ I = \frac{2}{\pi} \left( I_1 - I_2 \right) \]

where

\[ I_1 = \int \sin^{-1}(\sqrt{x}) \, dx \]
and
\[ I_2 = \int \cos^{-1}(\sqrt{x}) \, dx \]

### Step 1: Solve \(I_1\)

For \(I_1\), we will use the substitution \(u = \sqrt{x}\), which gives \(x = u^2\) and \(dx = 2u \, du\). Therefore:

\[ I_1 = \int \sin^{-1}(u) \cdot 2u \, du = 2 \int u \sin^{-1}(u) \, du \]

Using integration by parts, let \(v = \sin^{-1}(u)\) and \(dw = u \, du\). Then, \(dv = \frac{1}{\sqrt{1-u^2}} \, du\) and \(w = \frac{u^2}{2}\).

Applying integration by parts:

\[ \int u \sin^{-1}(u) \, du = \frac{u^2}{2} \sin^{-1}(u) - \int \frac{u^2}{2} \cdot \frac{1}{\sqrt{1-u^2}} \, du \]

Now we need to solve the remaining integral.

### Step 2: Solve \(I_2\)

For \(I_2\), we will also use the same substitution \(u = \sqrt{x}\):

\[ I_2 = \int \cos^{-1}(u) \cdot 2u \, du = 2 \int u \cos^{-1}(u) \, du \]

Using integration by parts again, let \(v = \cos^{-1}(u)\) and \(dw = u \, du\). Then, \(dv = -\frac{1}{\sqrt{1-u^2}} \, du\) and \(w = \frac{u^2}{2}\).

Applying integration by parts:

\[ \int u \cos^{-1}(u) \, du = \frac{u^2}{2} \cos^{-1}(u) + \int \frac{u^2}{2} \cdot \frac{1}{\sqrt{1-u^2}} \, du \]

### Step 3: Combine Results

Now we can combine the results of \(I_1\) and \(I_2\):

\[ I = \frac{2}{\pi} \left( 2 \left( \frac{u^2}{2} \sin^{-1}(u) - \int \frac{u^2}{2} \cdot \frac{1}{\sqrt{1-u^2}} \, du \right) - 2 \left( \frac{u^2}{2} \cos^{-1}(u) + \int \frac{u^2}{2} \cdot \frac{1}{\sqrt{1-u^2}} \, du \right) \right) \]

After simplification, we find that the integrals involving \(\frac{u^2}{2} \cdot \frac{1}{\sqrt{1-u^2}}\) will cancel out, leading us to a final expression.

### Final Result

After substituting back \(u = \sqrt{x}\), we can express the final result in terms of \(x\):

\[ I = \frac{2}{\pi} \left( \sin^{-1}(\sqrt{x}) - \cos^{-1}(\sqrt{x}) \right) + C \]