`(i) int(sinx)/(sqrt(4+ cos^(2) x)) dx " "(ii) int(x^(2))/(sqrt(9+x^(6)))dx`

Answer

Let's solve the two parts of the question step by step.

### Part (i):
**Integral to solve:**
\[ \int \frac{\sin x}{\sqrt{4 + \cos^2 x}} \, dx \]

**Step 1: Substitution**
Let \( \cos x = t \). Then, differentiating both sides gives:
\[ -\sin x \, dx = dt \quad \Rightarrow \quad \sin x \, dx = -dt \]

**Step 2: Rewrite the integral**
Substituting \( \sin x \, dx \) and \( \cos x \) into the integral:
\[ \int \frac{\sin x}{\sqrt{4 + \cos^2 x}} \, dx = \int \frac{-dt}{\sqrt{4 + t^2}} \]

**Step 3: Change the limits**
This becomes:
\[ -\int \frac{dt}{\sqrt{4 + t^2}} \]

**Step 4: Use the integral formula**
The integral \( \int \frac{dt}{\sqrt{a^2 + t^2}} = \log(t + \sqrt{a^2 + t^2}) + C \) where \( a = 2 \):
\[ -\log(t + \sqrt{4 + t^2}) + C \]

**Step 5: Substitute back for \( t \)**
Now substituting back \( t = \cos x \):
\[ -\log(\cos x + \sqrt{4 + \cos^2 x}) + C \]

### Final Answer for Part (i):
\[ -\log(\cos x + \sqrt{4 + \cos^2 x}) + C \]

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### Part (ii):
**Integral to solve:**
\[ \int \frac{x^2}{\sqrt{9 + x^6}} \, dx \]

**Step 1: Rewrite the expression**
Notice that \( x^6 = (x^3)^2 \) and \( 9 = 3^2 \):
\[ \int \frac{x^2}{\sqrt{3^2 + (x^3)^2}} \, dx \]

**Step 2: Substitution**
Let \( x^3 = t \). Then, differentiating gives:
\[ 3x^2 \, dx = dt \quad \Rightarrow \quad x^2 \, dx = \frac{dt}{3} \]

**Step 3: Rewrite the integral**
Substituting into the integral gives:
\[ \int \frac{x^2 \, dx}{\sqrt{9 + x^6}} = \int \frac{\frac{dt}{3}}{\sqrt{9 + t^2}} = \frac{1}{3} \int \frac{dt}{\sqrt{9 + t^2}} \]

**Step 4: Use the integral formula**
Using the formula \( \int \frac{dt}{\sqrt{a^2 + t^2}} = \log(t + \sqrt{a^2 + t^2}) + C \) where \( a = 3 \):
\[ \frac{1}{3} \log(t + \sqrt{9 + t^2}) + C \]

**Step 5: Substitute back for \( t \)**
Now substituting back \( t = x^3 \):
\[ \frac{1}{3} \log(x^3 + \sqrt{9 + x^6}) + C \]

### Final Answer for Part (ii):
\[ \frac{1}{3} \log(x^3 + \sqrt{9 + x^6}) + C \]

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