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I've learnt ibp, substitution.. but it seemed that i cannot just use those to solve the problem.. Please help me..

$$\int\limits_{-1}^{1} \sin(x^3) \mathrm{d}x$$

Botond
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SSARAH 98
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  • -1<x<1 ........ – SSARAH 98 Dec 06 '17 at 11:19
  • Do you mean the integral $\int_{-1}^1 \sin(x^3) dx$ (you can type that in yourself by placing \int_{-1}^1 \sin(x^3) dx inside a pair of $) – achille hui Dec 06 '17 at 11:27
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    This is a poor duplicate of the suggested duplicate. The key there is that one can evaluate this integral: but that's hard. The key that makes this easy is that it's an odd function integrated symmetrically across its axis, and thus is zero. – davidlowryduda Dec 06 '17 at 11:43
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    It would be better if you used a more informative title, as well as correct spelling and grammar within your post. Further, you mentio nthat this is no9-14, which I assume refers to a book? But you haven't said which book. – davidlowryduda Dec 06 '17 at 11:45

3 Answers3

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Both of the $f(x)=\sin(x)$ and the $f(x)=x^3$ are odd function, so they satisfy the following equation: $f(-x)=-f(x)$. $$\sin((-x)^3)=\sin(-x^3)=-\sin(x^3)$$ So $\sin(x^3)$ is also an odd function. The integral of the odd functions from $-a$ to $a$ is $0$: $$\int\limits_{-a}^{a} f(x) \mathrm{d}x=0$$ So: $$\int\limits_{-1}^{1} \sin(x^3) \mathrm{d}x=0$$ A little "proof": $$I=\int\limits_{0}^{a} f(x) \mathrm{d}x$$ Substitute $u=-x$, $\mathrm{d}u=-\mathrm{d}x$: $$I=\int\limits_{0}^{-a} f(-u) (-\mathrm{d}u)$$ We can change the order of limits: $$I=-\int\limits_{-a}^{0} f(-u) (-\mathrm{d}u)$$ $$I=\int\limits_{-a}^{0} f(-u) \mathrm{d}u$$ But because $f(-x)=-f(x)$: $$I=-\int\limits_{-a}^{0} f(u) \mathrm{d}u$$ Replacing $u$ with $x$, and multiplyng by $-1$: $$\int\limits_{-a}^{0} f(x) \mathrm{d}x=-I$$ So the $2$ equation for $I$: $$\int\limits_{-a}^{0} f(x) \mathrm{d}x=-I$$ $$\int\limits_{0}^{a} f(x) \mathrm{d}x=I$$ Adding the $2$ together: $$\int\limits_{-a}^{0} f(x) \mathrm{d}x+\int\limits_{0}^{a} f(x) \mathrm{d}x=I-I$$ $$\int\limits_{-a}^{a} f(x) \mathrm{d}x=0$$ Note: This method only works if both of the integrals exist, so $\int\limits_{-\infty}^{\infty} \sin(x) \mathrm{d}x \neq 0$, because neither the $\int\limits_{-\infty}^{0} \sin(x) \mathrm{d}x$ nor the $\int\limits_{0}^{\infty} \sin(x) \mathrm{d}x$ exist.

Botond
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Like Evaluate the integral $\int^{\frac{\pi}{2}}_0 \frac{\sin^3x}{\sin^3x+\cos^3x}dx$,

$$I=\int_{-a}^a\sin(x^{2n+1})dx=\int_{-a}^a\sin[\{a+(-a)-x\}^{2n+1}]dx=\cdots=-I$$

as $\sin(-u)=-\sin u$

lab bhattacharjee
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$$I=\int\limits_{-1}^{1} \sin(x^3) \mathrm{d}x \overset{u=-x}{=}-\int\limits_{-1}^{1} \sin(u^3) \mathrm{d}u = -I$$

Then $$2I=0\implies I=0$$

Guy Fsone
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