The hardest integral?

Calculus Level 3

∫ − e e sin ⁡ ( x ) cos ⁡ ( x ) 1 − x 2 + x 4 d x = ? \large \int_{-e}^{e} \frac{\sin(x)\cos (x)}{1-x^{2}+x^{4}}dx = ?

π − e π-e 1 0 e e

Alberto Caldera Morante by Alberto Caldera Morante , 20, Spain

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2 solutions

Chew-Seong Cheong
Apr 15, 2018

The integrand sin ⁡ x cos ⁡ x 1 − x 2 + x 4 \dfrac {\sin x \cos x}{1-x^2+x^4} is odd. Therefore, the integral ∫ − e e sin ⁡ x cos ⁡ x 1 − x 2 + x 4 d x = 0 \displaystyle \int_{-e}^e \frac {\sin x \cos x}{1-x^2+x^4} dx = \boxed{0} .

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if you look at the function we have to integrate f ( x ) = s e n ( x ) c o s ( x ) 1 − x 2 + x 4 f(x)=\frac{sen(x)cos (x)}{1-x^{2}+x^{4}} we see that f ( − x ) = − f ( x ) f(-x)=-f(x)

f ( − x ) = s e n ( − x ) c o s ( − x ) 1 − ( − x ) 2 + ( − x ) 4 = − s e n ( x ) c o s ( x ) 1 − x 2 + x 4 = − f ( x ) f(-x)=\frac{sen(-x)cos (-x)}{1-(-x)^{2}+(-x)^{4}}=-\frac{sen(x)cos (x)}{1-x^{2}+x^{4}}=-f(x)

and when f ( − x ) = − f ( x ) f(-x)=-f(x) , f ( x ) f(x) is an odd function

if f ( x ) f(x) is an odd function ∫ − a a f ( x ) d x = 0 \int_{-a}^{a} f(x)dx=0

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@Alberto Caldera Morante What do you mean by sen(x) ??

Aaghaz Mahajan - 3 years, 1 month ago

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sen x = sin x

Naren Bhandari - 3 years, 1 month ago

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