QJEEAdv15Ma07d6: JEE Advanced 2015 Mathematics - Class 12 JEEAdv15Ma07d6 Solution: Let for all Then the correct expression(s) is (are)

$f(x) = 7{\tan ^8}x + 7{\tan ^6}x - 3{\tan ^4}x - 3{\tan ^2}x\,\forall x \in \left( {{{ - \pi } \over 2},{\pi \over 2}} \right)$

$= 7{\tan ^6}x\,.\,{\sec ^2}x - 3{\tan ^2}x\,.\,{\sec ^2}x$

$= (7{\tan ^6}x - 3{\tan ^2}x)\,.\,{\sec ^2}x$

$\Rightarrow \int\limits_0^{\pi /4} {f(x)dx = \int\limits_0^{\pi /4} {(7{{\tan }^6}x - 3{{\tan }^2}x){{\sec }^2}dx = \int\limits_0^1 {(7 + 603{t^2})dt = [{t^7} - {t^3}]_0^1 = 0} } }$

Also, $I = \int\limits_0^{\pi /4} {xf(x)dx}$

$= \left| {x\,.\,\int {(7{{\tan }^6}x - 3{{\tan }^2}x){{\sec }^2}xdx} } \right|_0^{\pi /4} - \int\limits_0^{\pi /4} {1\,.\,\int {(7{{\tan }^6}x - 3{{\tan }^2}x){{\sec }^2}x\,dx} }$

$= \left| {x\,.\,({{\tan }^7}x - {{\tan }^3}x} \right|_0^{\pi /4} - \int\limits_0^{\pi /4} {({{\tan }^7}x - {{\tan }^3}x)dx}$

$= 0 - \int\limits_0^{\pi /4} {{{\tan }^3}x({{\tan }^4}x - 1)dx}$

$= - \int\limits_0^{\pi /4} {{{\tan }^3}x({{\tan }^2}x - 1)({{\sec }^2}x)dx}$

$= - \int\limits_0^1 {({t^5} - {t^3})dt}$

$= - \left[ {{{{t^6}} \over 6} - {{{t^4}} \over 4}} \right]_0^1 = \left[ {{1 \over 4} - {1 \over 6}} \right] = {1 \over {12}}$