Let f(x)=7tan8⁡x+7tan6⁡x−3tan4⁡x−3tan2⁡x for all x∈−π2,π2. Then the correct expression is

A
$\int_{0}^{π / 4} x f (x) d x = \frac{1}{12}$ and $\int_{0}^{π / 4} f (x) d x = 0$
B
$\int_{0}^{π / 4} f (x) d x = 2$
C
$\int_{0}^{π / 4} x f (x) d x = \frac{1}{6}$
D
$\int_{0}^{π / 4} f (x) d x = 1$

Solution:

We have,

$\begin{array}{l} f (x) = 7 \tan^{6} x (1 + \tan^{2} x) - 3 \tan^{2} x (1 + \tan^{2} x) \\ = (7 \tan^{6} x - 3 \tan^{2} x) \sec^{2} x \end{array}$

$∴ \int_{0}^{π / 4} f (x) d x = \int_{0}^{π / 4} (7 \tan^{6} x - 3 \tan^{2} x) \sec^{2} x d x$

$\Rightarrow \int_{0}^{π / 4} f (x) d x = \int_{0}^{1} (7 t^{6} - 3 t^{2}) d t,$ where $t = \tan x$ .

$\begin{array}{l} \int_{0}^{π / 4} x f (x) d x \\ = {[x (\tan^{7} x - \tan^{3} x)]}_{0}^{π / 4} - \int_{0}^{π / 4} (\tan^{7} x - \tan^{3} x) d x \\ = - \int_{0}^{π / 4} \tan^{3} x (\tan^{4} x - 1) d x \end{array}$

$\begin{array}{l} = - \int_{0}^{π / 4} \tan^{3} x (\tan^{2} x - 1) (\tan^{2} x + 1) d x \\ = - \int_{0}^{π / 4} \tan^{3} x (\tan^{2} x - 1) \sec^{2} x d x \end{array}$

$= - \int_{0}^{1} t^{3} (t^{2} - 1) d t,$ where $t = \tan x$

$= - {[\frac{t^{6}}{6} - \frac{t^{4}}{4}]}_{0}^{1} = \frac{1}{12}$

Let $f (x) = 7 \tan^{8} x + 7 \tan^{6} x - 3 \tan^{4} x - 3 \tan^{2} x$ for all $x \in (- \frac{π}{2}, \frac{π}{2}) .$ Then the correct expression is

Solution:

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