Evaluate :
(i) `int_(0)^(1)(3sqrt(x^(2))-4sqrt(x))/(sqrt(x))dx` , (ii) `int_(0)^(1)x cos(tan^(1)x)dx`

Answer

Let's solve the given integrals step by step.

### Part (i): Evaluate the integral

\[ \int_{0}^{1} \frac{3\sqrt{x^2} - 4\sqrt{x}}{\sqrt{x}} \, dx \]

**Step 1: Simplify the integrand**

We can simplify the integrand:

\[ \frac{3\sqrt{x^2} - 4\sqrt{x}}{\sqrt{x}} = \frac{3x - 4\sqrt{x}}{\sqrt{x}} = 3\frac{x}{\sqrt{x}} - 4\frac{\sqrt{x}}{\sqrt{x}} = 3\sqrt{x} - 4 \]

**Step 2: Rewrite the integral**

Now we can rewrite the integral:

\[ \int_{0}^{1} (3\sqrt{x} - 4) \, dx \]

**Step 3: Integrate term by term**

Now we can integrate each term separately:

\[ \int_{0}^{1} 3\sqrt{x} \, dx - \int_{0}^{1} 4 \, dx \]

The integral of \(3\sqrt{x}\) is:

\[ 3 \cdot \frac{2}{3} x^{3/2} = 2x^{3/2} \]

And the integral of \(4\) is:

\[ 4x \]

**Step 4: Evaluate the definite integrals**

Now we evaluate from 0 to 1:

\[ \left[ 2x^{3/2} \right]_{0}^{1} - \left[ 4x \right]_{0}^{1} \]

Calculating these:

1. For \(2x^{3/2}\):
- At \(x = 1\): \(2(1)^{3/2} = 2\)
- At \(x = 0\): \(2(0)^{3/2} = 0\)

Thus, \(2 - 0 = 2\).

2. For \(4x\):
- At \(x = 1\): \(4(1) = 4\)
- At \(x = 0\): \(4(0) = 0\)

Thus, \(4 - 0 = 4\).

Putting it all together:

\[ 2 - 4 = -2 \]

So, the answer for part (i) is:

\[ \int_{0}^{1} \frac{3\sqrt{x^2} - 4\sqrt{x}}{\sqrt{x}} \, dx = -2 \]

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### Part (ii): Evaluate the integral

\[ \int_{0}^{1} x \cos(\tan^{-1}(x)) \, dx \]

**Step 1: Substitute for \(\cos(\tan^{-1}(x))\)**

Let \(\theta = \tan^{-1}(x)\). Then:

\[ \tan(\theta) = x \implies \cos(\theta) = \frac{1}{\sqrt{1 + x^2}} \]

So, we can rewrite the integral:

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

**Step 2: Use substitution**

Let \(t = 1 + x^2\), then \(dt = 2x \, dx\) or \(dx = \frac{dt}{2x}\).

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

Now substituting:

\[ \int_{1}^{2} x \cdot \frac{1}{\sqrt{t}} \cdot \frac{dt}{2x} = \frac{1}{2} \int_{1}^{2} \frac{1}{\sqrt{t}} \, dt \]

**Step 3: Integrate**

The integral of \(\frac{1}{\sqrt{t}}\) is:

\[ 2\sqrt{t} \]

Thus:

\[ \frac{1}{2} \cdot \left[ 2\sqrt{t} \right]_{1}^{2} = \left[ \sqrt{t} \right]_{1}^{2} \]

**Step 4: Evaluate the definite integral**

Calculating:

\[ \sqrt{2} - \sqrt{1} = \sqrt{2} - 1 \]

So, the answer for part (ii) is:

\[ \int_{0}^{1} x \cos(\tan^{-1}(x)) \, dx = \sqrt{2} - 1 \]

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### Summary of Answers:
(i) \(-2\)
(ii) \(\sqrt{2} - 1\)

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