`(i) int(tan^(-1))/((1+x^(2)))dx" "(ii) int(1)/(sqrt(1-x^(2)) sin^(-1)x)dx`

Answer

Let's solve the given integrals step by step.

### (i) Evaluate the integral \( I_1 = \int \frac{\tan^{-1} x}{1 + x^2} \, dx \)

**Step 1: Substitution**
Let \( t = \tan^{-1} x \). Then, differentiate both sides with respect to \( x \):
\[ \frac{dt}{dx} = \frac{1}{1 + x^2} \]
This implies:
\[ dx = (1 + x^2) dt \]

**Step 2: Rewrite the integral**
Substituting \( t \) and \( dx \) into the integral:
\[ I_1 = \int t \cdot \frac{1}{1 + x^2} (1 + x^2) dt = \int t \, dt \]

**Step 3: Integrate**
Now, we can integrate:
\[ \int t \, dt = \frac{t^2}{2} + C_1 \]

**Step 4: Substitute back**
Replace \( t \) with \( \tan^{-1} x \):
\[ I_1 = \frac{(\tan^{-1} x)^2}{2} + C_1 \]

### Final Result for (i):
\[ I_1 = \frac{(\tan^{-1} x)^2}{2} + C_1 \]

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### (ii) Evaluate the integral \( I_2 = \int \frac{1}{\sqrt{1 - x^2}} \sin^{-1} x \, dx \)

**Step 1: Substitution**
Let \( t = \sin^{-1} x \). Then, differentiate both sides:
\[ \frac{dt}{dx} = \frac{1}{\sqrt{1 - x^2}} \]
This implies:
\[ dx = \sqrt{1 - x^2} \, dt \]

**Step 2: Rewrite the integral**
Substituting \( t \) and \( dx \) into the integral:
\[ I_2 = \int \frac{1}{\sqrt{1 - x^2}} t \cdot \sqrt{1 - x^2} \, dt = \int t \, dt \]

**Step 3: Integrate**
Now, we can integrate:
\[ \int t \, dt = \frac{t^2}{2} + C_2 \]

**Step 4: Substitute back**
Replace \( t \) with \( \sin^{-1} x \):
\[ I_2 = \frac{(\sin^{-1} x)^2}{2} + C_2 \]

### Final Result for (ii):
\[ I_2 = \frac{(\sin^{-1} x)^2}{2} + C_2 \]

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