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. ${\int }_0^{\frac{\pi }{4}}xf\left(x\right)dx=\frac{1}{12}$
• B. ${\int }_0^{\frac{\pi }{4}}f\left(x\right)dx=0$
• C. ${\int }_0^{\frac{\pi }{4}}xf\left(x\right)dx=\frac{1}{6}$
• D. ${\int }_0^{\frac{\pi }{4}}xf\left(x\right)dx=1$

Answer 1

Answer: (A), (B)

The correct options are
A ${\int }_0^{\frac{\pi }{4}}xf\left(x\right)dx=\frac{1}{12}$
B ${\int }_0^{\frac{\pi }{4}}f\left(x\right)dx=0$

$f\left(x\right)=7{\tan }^{8}x+7{\tan }^{6}x-3{\tan }^{4}x-3{\tan }^{2}x=(7{\tan }^{6}x-3{\tan }^{2}x){\sec }^{2}x$
$\therefore {\int }_0^{\frac{\pi }{4}}(7{\tan }^{6}x-3{\tan }^{2}x){\sec }^{2}xdx={[{\tan }^{7}x-{\tan }^{3}x]}_0^{\frac{\pi }{4}}=0$
And ${\int }_0^{\frac{\pi }{4}}x(7{\tan }^{6}x-3{\tan }^{2}x){\sec }^{2}xdx={\left[x\right({\tan }^{7}x-{\tan }^{3}x\left)\right]}_0^{\frac{\pi }{4}}-{\int }_0^{\frac{\pi }{4}}({\tan }^{7}x-{\tan }^{3}x)dx$
$={\int }_0^{\frac{\pi }{4}}{\tan }^{3}x(1-{\tan }^{4}x)dx={\int }_0^{\frac{\pi }{4}}{\tan }^{3}x(1-{\tan }^{2}x){\sec }^{2}xdx={[\frac{{\tan }^{4}x}{4}-\frac{{\tan }^{6}x}{6}]}_0^{\frac{\pi }{4}}=\frac{1}{12}$