Question

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

• A. $\underset{0}{\overset{\frac{\pi }{4}}{\int }}xf\left(x\right)dx=\frac{1}{12}$
• B. $\underset{0}{\overset{\frac{\pi }{4}}{\int }}f\left(x\right)dx=0$
• C. $\underset{0}{\overset{\frac{\pi }{4}}{\int }}xf\left(x\right)dx=\frac{1}{6}$
• D. $\underset{0}{\overset{\frac{\pi }{4}}{\int }}f\left(x\right)dx=1$

Answer 1

Answer: (A), (B)

$\underset{0}{\overset{\frac{\pi }{4}}{\int }}f\left(x\right)dx=0$

$\underset{0}{\overset{\frac{\pi }{4}}{\int }}f\left(x\right)dx=\underset{0}{\overset{\frac{\pi }{4}}{\int }}(7{\tan }^{6}x({\sec }^{2}x-1)+7{\tan }^{6}x-3{\tan }^{2}x({\sec }^{2}x-1)-3{\tan }^{2}x)dx$
$=\underset{0}{\overset{\frac{\pi }{4}}{\int }}(7{\tan }^{6}x.{\sec }^{2}x-3{\tan }^{2}x.{\sec }^{2}x)dx$
Let $\tan x=t\Rightarrow {\sec }^{2}xdx=dt$
$=\underset{0}{\overset{1}{\int }}(7t^6-3t^2)dt$
$\Rightarrow \underset{0}{\overset{\frac{\pi }{4}}{\int }}f\left(x\right)dx=[t^7{]}_0^1-[t^3{]}_0^1=1-1=0$
Also, $\underset{0}{\overset{\frac{\pi }{4}}{\int }}x.f\left(x\right)dx=\underset{0}{\overset{\frac{\pi }{4}}{\int }}(x.7{\tan }^{6}x.{\sec }^{2}x-\underset{0}{\overset{\frac{\pi }{4}}{\int }}x.3{\tan }^{2}x.{\sec }^{2}x)dx$
$=[x{\tan }^{7}x{]}_0^{\frac{\pi }{4}}-\underset{0}{\overset{\frac{\pi }{4}}{\int }}{\tan }^{7}xdx-[x{\tan }^{3}x{]}_0^{\frac{\pi }{4}}+\underset{0}{\overset{\frac{\pi }{4}}{\int }}{\tan }^{3}xdx$
$=\frac{\pi }{4}-\underset{0}{\overset{\frac{\pi }{4}}{\int }}{\tan }^{3}x({\tan }^{4}x-1)dx-\frac{\pi }{4}$
$=-\underset{0}{\overset{\frac{\pi }{4}}{\int }}{\tan }^{3}x({\tan }^{2}x-1){\sec }^{2}xdx$
Let $\tan x=t\Rightarrow {\sec }^{2}xdx=dt$
$=-\underset{0}{\overset{1}{\int }}(t^5-t^3)dt=-[\frac{1}{6}-\frac{1}{4}]=\frac{1}{12}$