If `inttan^(3)x sec^(3)xdx=((1)/(m))sec^(m)x-((1)/(n))sec^(n)x+c,` then `(m, n)-=`

Options

A. `(5, 3)`
B. `(3, 5)`
C. `((1)/(5),(1)/(3))`
D. `(4, 4)`

Answer

To solve the integral \( \int \tan^3 x \sec^3 x \, dx \) and express it in the form \( \frac{1}{m} \sec^m x - \frac{1}{n} \sec^n x + c \), we will follow these steps:

### Step 1: Rewrite the integrand
We start by rewriting \( \tan^3 x \) and \( \sec^3 x \):
\[ \tan^3 x = \tan x \tan^2 x = \tan x (\sec^2 x - 1) \]
\[ \sec^3 x = \sec x \sec^2 x \]
Thus, we can express the integral as:
\[ \int \tan^3 x \sec^3 x \, dx = \int \tan x (\sec^2 x - 1) \sec^3 x \, dx \]

### Step 2: Expand the integral
Now, we can expand the integral:
\[ \int \tan x \sec^3 x \sec^2 x \, dx - \int \tan x \sec^3 x \, dx \]

### Step 3: Use substitution
Let \( u = \sec x \), then \( du = \sec x \tan x \, dx \). Therefore, we have:
\[ \tan x \sec^3 x \, dx = u^3 \frac{du}{\sec x} = u^3 du \]
Thus, the integral becomes:
\[ \int u^3 \, du - \int u^3 \frac{du}{u} = \int u^4 \, du - \int u^2 \, du \]

### Step 4: Integrate
Now we integrate:
\[ \int u^4 \, du = \frac{u^5}{5} + C_1 \]
\[ \int u^2 \, du = \frac{u^3}{3} + C_2 \]
Putting it all together:
\[ \int \tan^3 x \sec^3 x \, dx = \frac{u^5}{5} - \frac{u^3}{3} + C \]

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

### Step 6: Identify \( m \) and \( n \)
From the expression, we can identify:
\[ m = 5, \quad n = 3 \]

### Final Answer
Thus, the values of \( (m, n) \) are:
\[ (m, n) = (5, 3) \]