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Provide solution for  RD Sharma maths Class 12 Chapter 19 Definite Integrals Exercise 19.4 (a) Question 9 textbook solution.

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Answer : 2

Given : \int_{\frac{-\pi}{4}}^{\frac{\pi}{4}} \frac{x^{11}-3 x^{9}+5 x^{7}-x^{5}+1}{\cos ^{2} x} d x

Hint : You must know about the concept of odd and even of f(x)

Solution : I=\int_{\frac{-\pi}{4}}^{\frac{\pi}{4}} \frac{x^{11}-3 x^{9}+5 x^{7}-x^{5}+1}{\cos ^{2} x} d x

\begin{aligned} &I=\int_{\frac{-\pi}{4}}^{\frac{\pi}{4}} \frac{x^{11}-3 x^{9}+5 x^{7}-x^{5}}{\cos ^{2} x}+\frac{1}{\cos ^{2} x} d x \\ &I=\int_{\frac{-\pi}{4}}^{\frac{\pi}{4}} \frac{x^{11}-3 x^{9}+5 x^{7}-x^{5}}{\cos ^{2} x}+\sec ^{2} x d x \end{aligned}

Now, \frac{x^{11}-3 x^{9}+5 x^{7}-x^{5}}{\cos ^{2} x} is odd and \sec^{2}x is even

\begin{aligned} &I=0+2 \int_{0}^{\frac{\pi}{4}} \sec ^{2} x d x \\ &{\left[\int_{-a}^{a} f(x) d x=2 \int_{0}^{a} f(x) d x, \text { even and } 0, \text { odd }\right]} \\ &I=2(\tan x)_{0}^{\frac{\pi}{4}} \\ &I=2 \end{aligned}

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