Evaluate: (i) `int(e^(sqrt(x))cos(e^(sqrt(x))))/(sqrt(x))\ dx` (ii) `int(cos^5x)/(sinx)\ dx`

Answer

Let's solve the integrals step by step.

### (i) Evaluate the integral:
\[ I_1 = \int \frac{e^{\sqrt{x}} \cos(e^{\sqrt{x}})}{\sqrt{x}} \, dx \]

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

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

**Step 3: Integrate**
The integral of \( \cos(t) \) is \( \sin(t) \):
\[ I_1 = 2 \sin(t) + C \]

**Step 4: Substitute back**
Substituting back \( t = e^{\sqrt{x}} \):
\[ I_1 = 2 \sin(e^{\sqrt{x}}) + C \]

### Final Answer for (i):
\[ I_1 = 2 \sin(e^{\sqrt{x}}) + C \]

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### (ii) Evaluate the integral:
\[ I_2 = \int \frac{\cos^5 x}{\sin x} \, dx \]

**Step 1: Rewrite the integral**
Rewrite \( \cos^5 x \) as \( \cos^4 x \cdot \cos x \):
\[ I_2 = \int \frac{\cos^4 x \cdot \cos x}{\sin x} \, dx \]

**Step 2: Use substitution**
Let \( t = \sin x \). Then, differentiate:
\[ \frac{dt}{dx} = \cos x \implies dt = \cos x \, dx \implies dx = \frac{dt}{\cos x} \]

**Step 3: Rewrite \( \cos^4 x \)**
Using the identity \( \cos^2 x = 1 - \sin^2 x \):
\[ \cos^4 x = (1 - t^2)^2 \]
Thus,
\[ I_2 = \int \frac{(1 - t^2)^2 \cdot \cos x}{t} \cdot \frac{dt}{\cos x} = \int \frac{(1 - t^2)^2}{t} \, dt \]

**Step 4: Expand and separate the integral**
Expanding \( (1 - t^2)^2 \):
\[ (1 - t^2)^2 = 1 - 2t^2 + t^4 \]
So,
\[ I_2 = \int \left( \frac{1}{t} - 2t + t^3 \right) dt \]

**Step 5: Integrate**
Now, integrate term by term:
\[ I_2 = \int \frac{1}{t} \, dt - 2 \int t \, dt + \int t^3 \, dt \]
This gives:
\[ I_2 = \ln |t| - t^2 + \frac{t^4}{4} + C \]

**Step 6: Substitute back**
Substituting \( t = \sin x \):
\[ I_2 = \ln |\sin x| - \sin^2 x + \frac{\sin^4 x}{4} + C \]

### Final Answer for (ii):
\[ I_2 = \ln |\sin x| - \sin^2 x + \frac{\sin^4 x}{4} + C \]

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