Let f(x) = 7 tan8 x + 7tan6x – 3tan4x – 3tan2x for all x \textrm{ } \in \textrm{ } \left(\right. - \textrm{ } \frac{\pi}{2} , \textrm{ }\textrm{ } \frac{\pi}{2} \left.\right). Then the correct expression(s) is(are)

Engineering

Mathematics

Properties of Definite Integral

Question

Let f(x) = 7 tan⁸ x + 7tan⁶x – 3tan⁴x – 3tan²x for all $x \in (- \frac{π}{2}, \frac{π}{2})$ . Then the correct expression(s) is(are)

$\int_{0}^{π / 4} x f (x) dx = \frac{1}{6}$

$\int_{0}^{π / 4} f (x) dx = 0$

$\int_{0}^{π / 4} f (x) dx = 1$

$\int_{0}^{π / 4} x f (x) dx = \frac{1}{12}$

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$\sqrt{f (x)}$ = 7 tan⁶x(tan²x + 1) – 3tan²x(tan²x + 1) = (7 tan⁶x – 3 tan²x) sec²x

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

Put tan x = t ⇒ sec²x dx = dt

$\int_{0}^{1} (7 t^{6} - 3 t^{2}) dt = 0$

$\int_{0}^{π / 4} x f (x) dx = \int_{0}^{π / 4} \underset{I}{\underset{⏟}{x}} \underset{II}{\underset{⏟}{(7 \tan^{6} x - 3 \tan^{2} x)}} \sec^{2} x dx$

$= {(x \cdot (\tan^{7} x - \tan^{3}))}_{0}^{π / 4} - \int_{0}^{π / 4} (\tan^{7} x - \tan^{3} x) dx$

$= 0 - \int_{0}^{π / 4} \tan^{3} x (\tan^{4} x - 1) dx$

$= - \int_{0}^{π / 4} \tan^{3} x (\tan^{2} x - 1) \sec^{2} x dx$

tan x = t

sec²x dx = dt

$= - \int_{0}^{1} t^{3} (t^{2} - 1) dt = - (\frac{1}{6} - \frac{1}{4}) = \frac{1}{12}$