`int(sin^3 2x)/(cos^5 2x)dx=`

Options

A. `tan^4x+c`
B. `tan 4x+c`
C. `tan^4 2x+x+c`
D. `(1)/(8)tan^4 2x+c`

Answer

To solve the integral \( \int \frac{\sin^3(2x)}{\cos^5(2x)} \, dx \), we can follow these steps:

### Step 1: Rewrite the integral
We start with the integral:
\[ I = \int \frac{\sin^3(2x)}{\cos^5(2x)} \, dx \]
We can rewrite \( \sin^3(2x) \) as \( \sin(2x) \cdot \sin^2(2x) \) and use the identity \( \sin^2(2x) = 1 - \cos^2(2x) \):
\[ I = \int \frac{\sin(2x)(1 - \cos^2(2x))}{\cos^5(2x)} \, dx \]
This simplifies to:
\[ I = \int \left( \frac{\sin(2x)}{\cos^5(2x)} - \frac{\sin(2x) \cos^2(2x)}{\cos^5(2x)} \right) \, dx \]
\[ I = \int \frac{\sin(2x)}{\cos^5(2x)} \, dx - \int \frac{\sin(2x)}{\cos^3(2x)} \, dx \]

### Step 2: Use substitution
Let \( t = \tan(2x) \). Then, we have:
\[ \frac{dt}{dx} = 2 \sec^2(2x) \quad \Rightarrow \quad dx = \frac{dt}{2 \sec^2(2x)} \]
Also, we know:
\[ \sin(2x) = \frac{t}{\sqrt{1+t^2}}, \quad \cos(2x) = \frac{1}{\sqrt{1+t^2}} \]
Thus, we can rewrite \( \sin(2x) \) and \( \cos(2x) \):
\[ \sin(2x) = t \cdot \frac{1}{\sqrt{1+t^2}}, \quad \cos(2x) = \frac{1}{\sqrt{1+t^2}} \]

### Step 3: Substitute into the integral
Now substituting these into the integral:
\[ I = \int \frac{t \cdot \frac{1}{\sqrt{1+t^2}}}{\left(\frac{1}{\sqrt{1+t^2}}\right)^5} \cdot \frac{dt}{2 \sec^2(2x)} \]
This simplifies to:
\[ I = \int \frac{t}{\frac{1}{(1+t^2)^{5/2}}} \cdot \frac{dt}{2 \cdot (1+t^2)} = \int \frac{t(1+t^2)^{5/2}}{2} \, dt \]

### Step 4: Integrate
Now we can integrate:
\[ I = \frac{1}{2} \int t^3 \, dt = \frac{1}{2} \cdot \frac{t^4}{4} + C = \frac{t^4}{8} + C \]

### Step 5: Substitute back
Substituting back \( t = \tan(2x) \):
\[ I = \frac{\tan^4(2x)}{8} + C \]

### Final Answer
Thus, the final answer is:
\[ \int \frac{\sin^3(2x)}{\cos^5(2x)} \, dx = \frac{\tan^4(2x)}{8} + C \]