Question

Evaluate each of the following integrals:

${\int }_{-\frac{\mathrm{\pi }}{4}}^{\frac{\mathrm{\pi }}{4}}\frac{x^{11}-3x^9+5x^7-x^5+1}{{\cos }^{2}x}dx$

Answer 1

$$\begin{gathered} \mathrm{Let}I={\int }_{-\frac{\mathrm{\pi }}{4}}^{\frac{\mathrm{\pi }}{4}}\frac{x^{11}-3x^9+5x^7-x^5+1}{{\cos }^{2}x}dx \\ ={\int }_{-\frac{\mathrm{\pi }}{4}}^{\frac{\mathrm{\pi }}{4}}\frac{x^{11}-3x^9+5x^7-x^5}{{\cos }^{2}x}dx+{\int }_{-\frac{\mathrm{\pi }}{4}}^{\frac{\mathrm{\pi }}{4}}\frac{1}{{\cos }^{2}x}dx \\ ={\int }_{-\frac{\mathrm{\pi }}{4}}^{\frac{\mathrm{\pi }}{4}}\frac{x^{11}-3x^9+5x^7-x^5}{{\cos }^{2}x}dx+{\int }_{-\frac{\mathrm{\pi }}{4}}^{\frac{\mathrm{\pi }}{4}}{\sec }^{2}xdx \\ =I_1+I_2 \end{gathered}$$

Now,

Consider $f\left(x\right)=\frac{x^{11}-3x^9+5x^7-x^5}{{\cos }^{2}x}$.

$\therefore f\left(-x\right)=\frac{{\left(-x\right)}^{11}-3{\left(-x\right)}^9+5{\left(-x\right)}^7-{\left(-x\right)}^5}{{\cos }^2\left(-x\right)}=\frac{-x^{11}+3x^9-5x^7+x^5}{{\cos }^{2}x}=-\frac{x^{11}-3x^9+5x^7-x^5}{{\cos }^{2}x}=-f\left(x\right)$

$\Rightarrow I_1={\int }_{-\frac{\mathrm{\pi }}{4}}^{\frac{\mathrm{\pi }}{4}}\frac{x^{11}-3x^9+5x^7-x^5}{{\cos }^{2}x}dx=0\left[{\int }_{-a}^{a}f\left(x\right)dx=\left\{\begin{array}{ll}2{\int }_0^{a}f\left(x\right)dx,& \mathrm{if}f\left(-x\right)=f\left(x\right)\\ 0,& \mathrm{if}f\left(-x\right)=-f\left(x\right)\end{array}\right.\right]$

Let $g\left(x\right)={\sec }^{2}x$.

$\therefore g\left(-x\right)={\sec }^2\left(-x\right)={\sec }^{2}x=g\left(x\right)$

$$\begin{gathered} \Rightarrow I_2={\int }_{-\frac{\mathrm{\pi }}{4}}^{\frac{\mathrm{\pi }}{4}}{\sec }^{2}xdx \\ =2{\int }_0^{\frac{\mathrm{\pi }}{4}}{\sec }^{2}xdx\left[{\int }_{-a}^{a}f\left(x\right)dx=\left\{\begin{array}{ll}2{\int }_0^{a}f\left(x\right)dx,& \mathrm{if}f\left(-x\right)=f\left(x\right)\\ 0,& \mathrm{if}f\left(-x\right)=-f\left(x\right)\end{array}\right.\right] \\ =2\times {\overline{)\tan x}}_0^{\frac{\mathrm{\pi }}{4}} \\ =2\left(\tan \frac{\mathrm{\pi }}{4}-\tan 0\right) \\ =2\times \left(1-0\right) \\ =2 \end{gathered}$$

$\therefore I=I_1+I_2=0+2=2$