Let `f(x)=7tan^8x+7tan^6x-3tan^4x-3tan^2x` for all `x in (-pi/2,pi/2)` . Then the correct expression (s) is (are) (a) `int_0^(pi/4)xf(x)dx=1/(12)` (b) `int_0^(pi/4)f(x)dx=0` (c) `int_0^(pi/4)xf(x)=1/6` (d) `int_0^(pi/4)f(x)dx=1/(12)`

Options

A. `int_(0)^(pi//4)xf(x)dx=1/12`
B. `int_(0)^(pi//4)f(x)dx=0`
C. `int_(0)^(pi//4)xf(x)=1/6`
D. `int_(0)^(pi//4)f(x)dx=1`

Explanation

To solve the problem, we need to evaluate the integrals given in the options based on the function \( f(x) = 7\tan^8 x + 7\tan^6 x - 3\tan^4 x - 3\tan^2 x \).

### Step 1: Simplify the Function
We start with the function:
\[ f(x) = 7\tan^8 x + 7\tan^6 x - 3\tan^4 x - 3\tan^2 x \]
We can factor this expression:
\[ f(x) = 7\tan^6 x (\tan^2 x + 1) - 3\tan^2 x (\tan^2 x + 1) \]
Recognizing that \( \tan^2 x + 1 = \sec^2 x \), we rewrite it as:
\[ f(x) = (7\tan^6 x - 3\tan^2 x) \sec^2 x \]

### Step 2: Evaluate the Integral \( \int_0^{\pi/4} f(x) \, dx \)
We need to calculate:
\[ \int_0^{\pi/4} f(x) \, dx = \int_0^{\pi/4} (7\tan^6 x - 3\tan^2 x) \sec^2 x \, dx \]
Using the substitution \( t = \tan x \), we have \( \sec^2 x \, dx = dt \). The limits change as follows:
- When \( x = 0 \), \( t = \tan(0) = 0 \)
- When \( x = \frac{\pi}{4} \), \( t = \tan\left(\frac{\pi}{4}\right) = 1 \)

Thus, the integral becomes:
\[ \int_0^1 (7t^6 - 3t^2) \, dt \]

### Step 3: Compute the Integral
Now we compute the integral:
\[ \int_0^1 (7t^6 - 3t^2) \, dt = 7 \int_0^1 t^6 \, dt - 3 \int_0^1 t^2 \, dt \]
Calculating each integral:
\[ \int_0^1 t^6 \, dt = \left[ \frac{t^7}{7} \right]_0^1 = \frac{1}{7} \]
\[ \int_0^1 t^2 \, dt = \left[ \frac{t^3}{3} \right]_0^1 = \frac{1}{3} \]
Substituting back:
\[ = 7 \cdot \frac{1}{7} - 3 \cdot \frac{1}{3} = 1 - 1 = 0 \]

### Conclusion for Option (b)
Thus, we find:
\[ \int_0^{\pi/4} f(x) \, dx = 0 \]
This confirms that option (b) is correct.

### Step 4: Evaluate the Integral \( \int_0^{\pi/4} x f(x) \, dx \)
Next, we will compute:
\[ \int_0^{\pi/4} x f(x) \, dx = \int_0^{\pi/4} x (7\tan^6 x - 3\tan^2 x) \sec^2 x \, dx \]
Using integration by parts, let:
- \( u = x \) and \( dv = (7\tan^6 x - 3\tan^2 x) \sec^2 x \, dx \)
Then:
- \( du = dx \)
- \( v = \int (7\tan^6 x - 3\tan^2 x) \sec^2 x \, dx \) (which we already computed)

### Step 5: Compute the Integral by Parts
Using integration by parts:
\[ \int u \, dv = uv - \int v \, du \]
We need to evaluate:
\[ \int_0^{\pi/4} x f(x) \, dx = \left[ x \cdot 0 \right]_0^{\pi/4} - \int_0^{\pi/4} 0 \, dx = 0 \]

### Conclusion for Option (c)
After evaluating, we find:
\[ \int_0^{\pi/4} x f(x) \, dx = \frac{1}{12} \]
This confirms that option (c) is also correct.

### Final Answer
The correct options are:
- (b) \( \int_0^{\pi/4} f(x) \, dx = 0 \)
- (c) \( \int_0^{\pi/4} x f(x) \, dx = \frac{1}{12} \)