The integral `I=int sec^(3)x tan^(3)xdx` is equal to (where, C is the constant of integration)

Options

A. `sec^(5)x-sec^(3)x+C`
B. `(sec^(5)x)/(5)-sec^(3)x+C`
C. `(sec^(5)x)/(5)-(sec^(3)x)/(3)+C`
D. `(sec^(5)x)/(5)-tan^(-1)x+C`

Answer

To solve the integral \( I = \int \sec^3 x \tan^3 x \, dx \), we can use substitution and integration techniques. Here’s a step-by-step solution:

### Step 1: Substitution
Let \( t = \sec x \). Then, the derivative of \( t \) with respect to \( x \) is:
\[ dt = \sec x \tan x \, dx \quad \Rightarrow \quad dx = \frac{dt}{\sec x \tan x} = \frac{dt}{t \sqrt{t^2 - 1}} \]
Here, we used the identity \( \tan^2 x = \sec^2 x - 1 \).

### Step 2: Rewrite the Integral
Now, we can rewrite the integral in terms of \( t \):
\[ I = \int \sec^3 x \tan^3 x \, dx = \int t^3 (t^2 - 1) \frac{dt}{t \sqrt{t^2 - 1}} \]
This simplifies to:
\[ I = \int t^2 (t^2 - 1) \, dt \]

### Step 3: Expand the Integral
Now, expand the integrand:
\[ I = \int (t^4 - t^2) \, dt \]

### Step 4: Integrate
Now, integrate term by term:
\[ I = \int t^4 \, dt - \int t^2 \, dt = \frac{t^5}{5} - \frac{t^3}{3} + C \]

### Step 5: Substitute Back
Now substitute back \( t = \sec x \):
\[ I = \frac{\sec^5 x}{5} - \frac{\sec^3 x}{3} + C \]

### Final Answer
Thus, the integral \( I \) is:
\[ I = \frac{\sec^5 x}{5} - \frac{\sec^3 x}{3} + C \]

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