Question

If $f(x) = 7{\tan ^8}x + 7{\tan ^6}x - 3\tan {^4}x - 3{\tan ^2}x$for all $x \in ( - \dfrac{\pi }{2},\dfrac{\pi }{2})$, then the correct expression(s) is/are
A.$\int\limits_0^{\dfrac{\pi }{4}} {xf(x)dx = \dfrac{1}{{12}}}$
B.$\int\limits_0^{\dfrac{\pi }{4}} {f(x)dx = 0}$
C.$\int\limits_0^{\dfrac{\pi }{4}} {xf(x)dx = \dfrac{1}{6}}$
D.$\int\limits_0^{\dfrac{\pi }{4}} {xf(x)dx = 1}$

Answer 1

Use the trigonometric formulas as $1 + {\tan ^2}x = {\sec ^2}x$, also if possible take the terms common and simplify them as much as possible. Hence use the substitution method for integration as $\tan x = t,{\sec ^2}xdx = dt$and also if required apply the rule of integration by-parts as$\int {yf(x).dx = y\int {f(x)} - \int {\dfrac{{dy}}{{dx}}(} } \int {f(x)} ).dx$hence from this concept we can get our required answer.

Complete step-by-step answer:
As the given function is as $f(x) = 7{\tan ^8}x + 7{\tan ^6}x - 3\tan {^4}x - 3{\tan ^2}x$for all $x \in ( - \dfrac{\pi }{2},\dfrac{\pi }{2})$
Firstly, simplify the $f\left( x \right)$, as per using various trigonometric formulas, so,
$\Rightarrow 7{\tan ^8}x + 7{\tan ^6}x - 3\tan {^4}x - 3{\tan ^2}x$
Taking $7{\tan ^6}x$and $3{\tan ^2}x$common from first two terms and last two terms respectively, we get,
$\Rightarrow 7{\tan ^6}x(1 + {\tan ^2}x) - 3{\tan ^2}x(1 + {\tan ^2}x)$
Now replace $1 + {\tan ^2}x = {\sec ^2}x$in the above equation as,
$\Rightarrow 7{\tan ^6}x({\sec ^2}x) - 3{\tan ^2}x({\sec ^2}x)$
Now taking ${\sec ^2}x$ as common we get,
$\Rightarrow (7{\tan ^6}x - 3{\tan ^2}x)({\sec ^2}x)$
Thus, now firstly calculate the integration of,
$\int\limits_0^{\dfrac{\pi }{4}} {f(x)dx}$,
On substituting the value of $f\left( x \right)$we get,
$\int\limits_0^{\dfrac{\pi }{4}} {(7{{\tan }^6}x - 3{{\tan }^2}x)({{\sec }^2}x)dx}$
Let us use the substitution method to calculate the integral of above equation as $\tan x = t,{\sec ^2}xdx = dt$
Hence the limit will now vary as $\tan 0 = 0,\tan \dfrac{\pi }{4} = 1$,
So the integration can be given as,
$= \int\limits_0^1 {(7{t^6} - 3{t^2})dt}$
On integrating we get,
$= 7(\dfrac{{{t^7}}}{7})|_0^1 - 3(\dfrac{{{t^3}}}{3})|_0^1$
On simplification we get,
$= {t^7} - {t^3}|_0^1$
On applying limits we get,
$= 1 - 0 - (1 - 0)$
On simplification we get,

$$= 1 - 1 \\\ = 0 \\\$$

Hence, $\int\limits_0^{\dfrac{\pi }{4}} {f(x)dx} = 0$
So, option (B) will be our correct answer.
Now, check for the integration of $\int\limits_0^{\dfrac{\pi }{4}} {xf(x)dx}$, and use the concept of by- parts for this so it will be,
$I = \int\limits_0^{\dfrac{\pi }{4}} {x(7{{\tan }^6}x - 3{{\tan }^2}x)({{\sec }^2}x)dx}$
As we know that the integration, of known quantity is as,
$\int {f(x)dx} = \int {(7{{\tan }^6}x - 3{{\tan }^2}x)({{\sec }^2}x)dx} = {\tan ^7}x - {\tan ^3}x + c$
So also use this in calculating the integration of $\int\limits_0^{\dfrac{\pi }{4}} {xf(x)dx}$,
On applying by-parts, which is $\int {uvdx = u\int {vdx - \int {\dfrac{{du}}{{dx}}\left( {\int v dx} \right)dx} } }$, we get,
$\int {x((7{{\tan }^6}x - 3{{\tan }^2}x)({{\sec }^2}x)dx).dx = x\int\limits_0^{\dfrac{\pi }{4}} {(7{{\tan }^6}x - 3{{\tan }^2}x)({{\sec }^2}x)dx} - \int {\dfrac{{dx}}{{dx}}(} } \int\limits_0^{\dfrac{\pi }{4}} {(7{{\tan }^6}x - 3{{\tan }^2}x)({{\sec }^2}x)dx} ).dx$
Hence after applying by parts directly we can place the value of known integration terms as,
$= [x({\tan ^7}x - {\tan ^3}x)]|_0^{\dfrac{\pi }{4}} - \int\limits_0^{\dfrac{\pi }{4}} {1(} {\tan ^7}x - {\tan ^3}x).dx$
Hence now calculate the equation of second part of the above equation as,
$I' = - \int\limits_0^{\dfrac{\pi }{4}} {1(} {\tan ^7}x - {\tan ^3}x).dx$
Taking $- {\tan ^3}x$, common we get,
\Rightarrow $$$$I' = \int\limits_0^{\dfrac{\pi }{4}} ( {\tan ^3}x)(1 - {\tan ^4}x).dx
Using, ${a^2} - {b^2} = (a + b)(a - b)$, we get,
\Rightarrow $$$$I' = \int\limits_0^{\dfrac{\pi }{4}} ( {\tan ^3}x)(1 - {\tan ^2}x)(1 + {\tan ^2}x).dx
Now using $1 + {\tan ^2}x = {\sec ^2}x$, we get,
\Rightarrow $$$$I' = \int\limits_0^{\dfrac{\pi }{4}} ( {\tan ^3}x)(1 - {\tan ^2}x)({\sec ^2}x).dx
Now, use the substitution method to solve the above integration as $\tan x = t,{\sec ^2}xdx = dt$
Hence the limit will now vary as $\tan 0 = 0,\tan \dfrac{\pi }{4} = 1$,
So, the integral can be given as
\Rightarrow $$$$I' = \int\limits_0^1 ( {\operatorname{t} ^3})(1 - {\operatorname{t} ^2})dt
On expanding we get,
$\Rightarrow I' = \int\limits_0^1 ( {\operatorname{t} ^3} - {\operatorname{t} ^5})dt$
On integration we get,
$\Rightarrow I' = \dfrac{{{\operatorname{t} ^4}}}{4}|_0^1 - \dfrac{{{\operatorname{t} ^6}}}{6}|_0^1$
On applying limits we get,
$\Rightarrow I' = \dfrac{1}{4} - \dfrac{1}{6}$
On taking LCM and simplification we get,

$$\Rightarrow I' = \dfrac{{6 - 4}}{{24}} \\\ \Rightarrow I' = \dfrac{2}{{24}} \\\ \Rightarrow I' = \dfrac{1}{{12}} \\\$$

Hence, so now we directly have to put all the values of various integrals in
$\int {x((7{{\tan }^6}x - 3{{\tan }^2}x)({{\sec }^2}x)dx).dx = x\int\limits_0^{\dfrac{\pi }{4}} {(7{{\tan }^6}x - 3{{\tan }^2}x)({{\sec }^2}x)dx} - \int {\dfrac{{dx}}{{dx}}(} } \int\limits_0^{\dfrac{\pi }{4}} {(7{{\tan }^6}x - 3{{\tan }^2}x)({{\sec }^2}x)dx} ).dx$
Thus,
$I = [x({\tan ^7}x - {\tan ^3}x)]|_0^{\dfrac{\pi }{4}} - \int\limits_0^{\dfrac{\pi }{4}} {1(} {\tan ^7}x - {\tan ^3}x).dx$
On substituting the values we get,

$$\Rightarrow I = \dfrac{\pi }{4}((1 - 1) - 0) + \dfrac{1}{{12}} \\\ \Rightarrow I = \dfrac{1}{{12}} \\\$$

Hence, option (A) and option (B) both are correct answers.

Note: In calculus, and more generally in mathematical analysis, integration by parts or partial integration is a process that finds the integral of a product of functions in terms of the integral of the product of their derivative and antiderivative.
Calculate and apply the property of integration as substitution and by parts rule carefully, also change the limit value also changing the substitution.