Evaluate the following integrals: ^ 4 _ - 4 x^ 11 -3x^9+5x^7-x^5+1 ^2x dx

Let I= ^ 4 _ - 4 x^ 11 -3x^9+5x^7-x^5+1 ^2x dx= ^ 4 _ - 4 x^ 11 -3x^9+5x^7-x^5 ^2x dx+ ^ 4 _ - 4 1 ^2x dx = ^ 4 _ - 4 x^ 11 -3x^9+5x^7-x^5 ^2x dx+ ^ 4 _ - 4 ^2x dx =I_1+I_2 Now, Consider, f(x)= x^ 11 -3x^9+5x^7-x^5 ^2x dx f(-x)= (-x)^ 11 -3(-x)^9+5(-x)^7-(-x)^5 ^2(-x) = -x^ 11 +3x^9-5x^7+x^5 ^2x =- x^ 11 -3x^9+5x^7-x^5 ^2x =f(x) _1= ^ 4 _ - 4 x^ 11 -3x^9+5x^7-x^5 ^2x dx=0 ^ a _0f(x)dx= ^ a _0f(x)dx,& if f(-x)=f(x) 0,& if f(-x)=-f(x) g(x)= ^2x Let g(-x)= ^2(-x)= ^2x=g(x) _1= ^ 4 _ - 4 ^2x dx =2 ^ 4 _0 ^2x dx ^ a _0f(x)dx= ^ a _0f(x)dx,& if f(-x)=f(x) 0,& if f(-x)=-f(x) =2 [ ]^ 4 _0 =2 ( 4 - 0 ) =2 (1-0) =2 I=I_1+I_2 I=0+2 I=2

Evaluate the following integrals: ^ 4 _ - 4 x^ 11 -3x^9+5x^7-x^5+…

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Question

Evaluate the following integrals:$\int\limits^{\frac{\pi}{4}}_{-\frac{\pi}{4}}\frac{\text{x}^{11}-3\text{x}^9+5\text{x}^7-\text{x}^5+1}{\cos^2\text{x}}\text{ dx}$

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Answer

Let $\text{I}=\int\limits^{\frac{\pi}{4}}_{-\frac{\pi}{4}}\frac{\text{x}^{11}-3\text{x}^9+5\text{x}^7-\text{x}^5+1}{\cos^2\text{x}}\text{ dx}$$=\int\limits^{\frac{\pi}{4}}_{-\frac{\pi}{4}}\frac{\text{x}^{11}-3\text{x}^9+5\text{x}^7-\text{x}^5}{\cos^2\text{x}}\text{ dx}+\int\limits^{\frac{\pi}{4}}_{-\frac{\pi}{4}}\frac{1}{\cos^2\text{x}}\text{ dx}$
$=\int\limits^{\frac{\pi}{4}}_{-\frac{\pi}{4}}\frac{\text{x}^{11}-3\text{x}^9+5\text{x}^7-\text{x}^5}{\cos^2\text{x}}\text{ dx}+\int\limits^{\frac{\pi}{4}}_{-\frac{\pi}{4}}\sec^2\text{x}\text{ dx}$
$=\text{I}_1+\text{I}_2$
Now,
Consider, $\text{f(x)}=\frac{\text{x}^{11}-3\text{x}^9+5\text{x}^7-\text{x}^5}{\cos^2\text{x}}\text{ dx}$
$\therefore\ \text{f}(-\text{x})=\frac{(-\text{x})^{11}-3(-\text{x})^9+5(-\text{x})^7-(-\text{x})^5}{\cos^2(-\text{x})}$
$=\frac{-\text{x}^{11}+3\text{x}^9-5\text{x}^7+\text{x}^5}{\cos^2\text{x}}$
$=-\frac{\text{x}^{11}-3\text{x}^9+5\text{x}^7-\text{x}^5}{\cos^2\text{x}}=\text{f(x)}$
$\Rightarrow\text{I}_1=\int\limits^{\frac{\pi}{4}}_{-\frac{\pi}{4}}\frac{\text{x}^{11}-3\text{x}^9+5\text{x}^7-\text{x}^5}{\cos^2\text{x}}\text{ dx}=0$ $\begin{bmatrix}\int\limits^{\text{a}}_0\text{f(x)}\text{dx}=\begin{cases}\int\limits^{\text{a}}_0\text{f(x)}\text{dx},&\text{ if }\text{ f}(-\text{x})=\text{f(x)}\\0,&\text{ if }\text{ f}(-\text{x})=-\text{f(x)}\end{cases}\end{bmatrix}$
$\text{g(x)}=\sec^2\text{x}$
Let $\text{g}(-\text{x})=\sec^2(-\text{x})=\sec^2\text{x}=\text{g}(\text{x})$
$\Rightarrow\text{I}_1=\int\limits^{\frac{\pi}{4}}_{-\frac{\pi}{4}}\sec^2\text{x dx}$
$=2\int\limits^{\frac{\pi}{4}}_0\sec^2\text{x dx}$ $\begin{bmatrix}\int\limits^{\text{a}}_0\text{f(x)}\text{dx}=\begin{cases}\int\limits^{\text{a}}_0\text{f(x)}\text{dx},&\text{ if }\text{ f}(-\text{x})=\text{f(x)}\\0,&\text{ if }\text{ f}(-\text{x})=-\text{f(x)}\end{cases}\end{bmatrix}$
$=2\times\Big[\tan\text{x}\Big]^{\frac{\pi}{4}}_0$
$=2\Big(\tan\frac{\pi}{4}-\tan0\Big)$
$=2\times(1-0)$
$=2$
$\therefore\ \text{I}=\text{I}_1+\text{I}_2$
$\text{I}=0+2$
$\text{I}=2$