Question

Evaluate each of the following integral:

\[\int_{- \frac{\pi}{4}}^\frac{\pi}{4} \frac{x^{11} - 3 x^9 + 5 x^7 - x^5 + 1}{\cos^2 x}dx\]

Answer 1

\[Let I = \int_{- \frac{\pi}{4}}^\frac{\pi}{4} \frac{x^{11} - 3 x^9 + 5 x^7 - x^5 + 1}{\cos^2 x}dx\]
\[ = \int_{- \frac{\pi}{4}}^\frac{\pi}{4} \frac{x^{11} - 3 x^9 + 5 x^7 - x^5}{\cos^2 x}dx + \int_{- \frac{\pi}{4}}^\frac{\pi}{4} \frac{1}{\cos^2 x}dx\]
\[ = \int_{- \frac{\pi}{4}}^\frac{\pi}{4} \frac{x^{11} - 3 x^9 + 5 x^7 - x^5}{\cos^2 x}dx + \int_{- \frac{\pi}{4}}^\frac{\pi}{4} \sec^2 xdx\]
\[ = I_1 + I_2\]

Now,

Consider

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

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

\[\Rightarrow I_1 = \int_{- \frac{\pi}{4}}^\frac{\pi}{4} \frac{x^{11} - 3 x^9 + 5 x^7 - x^5}{\cos^2 x}dx = 0 ..................\left[ \int_{- a}^a f\left( x \right)dx = \begin{cases}2 \int_0^a f\left( x \right)dx, & \text{if }f\left( - x \right) = f\left( x \right) \\ 0, & \text{if }f\left( - x \right) = - f\left( x \right)\end{cases} \right]\]

Let

\[g\left( x \right) = \sec^2 x\]

\[\therefore g\left( - x \right) = \sec^2 \left( - x \right) = \sec^2 x = g\left( x \right)\]

\[\Rightarrow I_2 = \int_{- \frac{\pi}{4}}^\frac{\pi}{4} \sec^2 xdx\]
\[ = 2 \int_0^\frac{\pi}{4} \sec^2 xdx ...................\left[ \int_{- a}^a f\left( x \right)dx = \begin{cases}2 \int_0^a f\left( x \right)dx, & \text{if }f\left( - x \right) = f\left( x \right) \\ 0, & \text{if }f\left( - x \right) = - f\left( x \right)\end{cases} \right]\]
\[ = 2 \times \left.\tan x\right|_0^\frac{\pi}{4} \]
\[ = 2\left( \tan\frac{\pi}{4} - \tan0 \right)\]
\[ = 2 \times \left( 1 - 0 \right)\]

\[ = 2\]

\[\therefore I = I_1 + I_2 = 0 + 2 = 2\]