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Definite Integrals — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsDefinite Integrals5 Marks
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 $\text{g}(-\text{x})=\sec^2(-\text{x})=\sec^2\text{x}=\text{g}(\text{x})$
$\Rightarrow\text{I}_1=\int\limits^{\frac{\pi}{4}}_{-\frac{\pi}{4}}\sec^2\text{x dx}$
$=2\int\limits^{\frac{\pi}{4}}_0\sec^2\text{x dx}$ $\begin{bmatrix}\int\limits^{\text{a}}_0\text{f(x)}\text{dx}=\begin{cases}\int\limits^{\text{a}}_0\text{f(x)}\text{dx},&\text{ if }\text{ f}(-\text{x})=\text{f(x)}\\0,&\text{ if }\text{ f}(-\text{x})=-\text{f(x)}\end{cases}\end{bmatrix}$
$=2\times\Big[\tan\text{x}\Big]^{\frac{\pi}{4}}_0$
$=2\Big(\tan\frac{\pi}{4}-\tan0\Big)$
$=2\times(1-0)$
$=2$
$\therefore\ \text{I}=\text{I}_1+\text{I}_2$
$\text{I}=0+2$
$\text{I}=2$

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