If $$x2f(x)+f(1/x)=0$$ for all x∈R−$${0}$$ then∫$$cos$$⁡θ$$sec$$⁡θ $$f(x)dx=$$

$$I=$$∫$$cos$$⁡θ$$sec$$⁡θ $$f(x)dx=$$−∫$$cos$$⁡θ$$sec$$⁡θ $$1x2f1xdxPut$$⁡$$1x=t=$$∫$$sec$$⁡θ$$cos$$⁡θ $$f(t)dt=$$−∫$$cos$$⁡θ$$sec$$⁡θ $$f(t)dt=$$−IHence $$2I=0$$⇒$$I=0$$

Evaluation of definite integrals

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Question

If $x^{2} f (x) + f (1 / x) = 0$ for all $x \in R - {0}$ then
$\int_{\cos θ}^{\sec θ} f (x) d x =$

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Solution

$\begin{array}{l} I = \int_{\cos θ}^{\sec θ} f (x) d x = - \int_{\cos θ}^{\sec θ} \frac{1}{x^{2}} f (\frac{1}{x}) d x (Put \frac{1}{x} = t) \\ = \int_{\sec θ}^{\cos θ} f (t) d t = - \int_{\cos θ}^{\sec θ} f (t) d t = - I \end{array}$
Hence $2 I = 0 \Rightarrow I = 0$

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