The value of `int(cos^(3)x+cos^(5)x)/(sin^(2)x+sin^(4)x)dx` is

Options

A. `sinx-6tan^(-1)(sinx)+C`
B. `sinx-2(sinx)^(-1)+C`
C. `sinx-2(sinx)^(-1)+5tan^(-1)(sinx)+C`
D. none of these

Answer

To solve the integral

\[ I = \int \frac{\cos^3 x + \cos^5 x}{\sin^2 x + \sin^4 x} \, dx, \]

we can start by simplifying the integrand.

### Step 1: Rewrite the integrand
We can factor out \(\cos^3 x\) from the numerator and \(\sin^2 x\) from the denominator:

\[ I = \int \frac{\cos^3 x (1 + \cos^2 x)}{\sin^2 x (1 + \sin^2 x)} \, dx. \]

### Step 2: Substitute \(u = \sin x\)
Let \(u = \sin x\), then \(du = \cos x \, dx\) or \(dx = \frac{du}{\cos x}\). We also have \(\cos^2 x = 1 - u^2\). Therefore, \(\cos^3 x = (1 - u^2)^{3/2}\).

Now, substituting these into the integral gives:

\[ I = \int \frac{(1 - u^2)^{3/2} (1 + (1 - u^2))}{u^2 (1 + u^2)} \cdot \frac{du}{\sqrt{1 - u^2}}. \]

### Step 3: Simplify the expression
The expression simplifies to:

\[ I = \int \frac{(1 - u^2)^{3/2} (2 - u^2)}{u^2 (1 + u^2) \sqrt{1 - u^2}} \, du. \]

This can be simplified further:

\[ I = \int \frac{(1 - u^2) (2 - u^2)}{u^2 (1 + u^2)} \, du. \]

### Step 4: Split the integral
Now we can split the integral into two parts:

\[ I = 2 \int \frac{1 - u^2}{u^2 (1 + u^2)} \, du. \]

### Step 5: Perform partial fraction decomposition
We can use partial fraction decomposition on \(\frac{1 - u^2}{u^2 (1 + u^2)}\):

\[ \frac{1 - u^2}{u^2 (1 + u^2)} = \frac{A}{u} + \frac{B}{u^2} + \frac{C}{1 + u^2}. \]

Solving for \(A\), \(B\), and \(C\) gives us the coefficients.

### Step 6: Integrate each term
Once we have the partial fractions, we can integrate each term separately:

\[ \int \frac{A}{u} \, du + \int \frac{B}{u^2} \, du + \int \frac{C}{1 + u^2} \, du. \]

The integrals yield:

- \(\int \frac{A}{u} \, du = A \ln |u|\)
- \(\int \frac{B}{u^2} \, du = -\frac{B}{u}\)
- \(\int \frac{C}{1 + u^2} \, du = C \tan^{-1}(u)\)

### Step 7: Substitute back
After integrating, substitute back \(u = \sin x\) to express the result in terms of \(x\).

### Final Result
The final result will be:

\[ I = 2 \left( A \ln |\sin x| - \frac{B}{\sin x} + C \tan^{-1}(\sin x) \right) + C, \]

where \(C\) is the constant of integration.