Reduction formulas can be used to compute integrals of higher power of `sinx,cosx,tanx` etc.
`inttan^(6)xdx=(1)/(5)tan^(5)x+Atan^(3)x+tanx-x+C` then

Options

A. `A=(1)/(3)`
B. `A=(2)/(3)`
C. `A=-(2)/(3)`
D. `A=-(1)/(3)`

Answer

To solve the integral \( \int \tan^6 x \, dx \) and find the value of \( A \) in the expression

\[ \int \tan^6 x \, dx = \frac{1}{5} \tan^5 x + A \tan^3 x + \tan x - x + C, \]

we can follow these steps:

### Step 1: Break down the integral
We start by rewriting \( \tan^6 x \) as \( \tan^4 x \cdot \tan^2 x \). This gives us:

\[ \int \tan^6 x \, dx = \int \tan^4 x \cdot \tan^2 x \, dx. \]

### Step 2: Substitute \( \tan^2 x \)
Using the identity \( \tan^2 x = \sec^2 x - 1 \), we can express \( \tan^2 x \) in terms of \( \sec^2 x \):

\[ \tan^6 x = \tan^4 x (\sec^2 x - 1). \]

Substituting this into the integral, we have:

\[ \int \tan^6 x \, dx = \int \tan^4 x \sec^2 x \, dx - \int \tan^4 x \, dx. \]

### Step 3: Solve the first integral
For the first integral, we can use the substitution \( t = \tan x \), which gives us \( dt = \sec^2 x \, dx \) or \( dx = \frac{dt}{\sec^2 x} \). Thus, we rewrite the integral:

\[ \int \tan^4 x \sec^2 x \, dx = \int t^4 \, dt = \frac{t^5}{5} + C = \frac{\tan^5 x}{5} + C. \]

### Step 4: Solve the second integral
For the second integral, we again use the substitution \( t = \tan x \):

\[ \int \tan^4 x \, dx = \int t^4 \cdot \frac{dt}{\sec^2 x} = \int t^4 \cdot \frac{dt}{1 + t^2}. \]

This integral can be solved using polynomial long division or other techniques, but for our purposes, we can express it as:

\[ \int t^4 \, dt = \frac{t^5}{5} - \int t^2 \, dt = \frac{t^5}{5} - \frac{t^3}{3} + C = \frac{\tan^5 x}{5} - \frac{\tan^3 x}{3} + C. \]

### Step 5: Combine results
Now we can combine the results of both integrals:

\[ \int \tan^6 x \, dx = \left( \frac{\tan^5 x}{5} \right) - \left( \frac{\tan^5 x}{5} - \frac{\tan^3 x}{3} \right) + \tan x - x + C. \]

This simplifies to:

\[ \int \tan^6 x \, dx = \frac{1}{5} \tan^5 x + \frac{1}{3} \tan^3 x + \tan x - x + C. \]

### Step 6: Identify \( A \)
Comparing with the given expression:

\[ \int \tan^6 x \, dx = \frac{1}{5} \tan^5 x + A \tan^3 x + \tan x - x + C, \]

we find that \( A = \frac{1}{3} \).

### Final Answer
Thus, the value of \( A \) is:

\[ A = \frac{1}{3}. \]