i've doubts about this integration

[logistic_guy]

here is the question

Evaluate the definite integral \(\displaystyle \int_{0}^{1} \sin x^3 \ dx\) to the indicated accuracy.


my attemps

is it possible to solve integration to 10 decimal places?

my first attemb to use the trigonometric identity \(\displaystyle \sin^3 x = \sin x^3\)

when i use \(\displaystyle x = 0.1, 0.2, 0.3\), they give different values but when i use \(\displaystyle x = 0, 1\) they give the same values. \(\displaystyle 0\) and \(\displaystyle 1\) are the integration limit and their values fit the trigonometric identity

if i solve \(\displaystyle \int_{0}^{1} \sin^3 x \ dx\) how to compare it with \(\displaystyle \int_{0}^{1} \sin x^3 \ dx\) when i don't know the result of the original integration?

 

[fresh_42]

[math] (\sin x)^3=\sin^3 x \neq \sin x^3=\sin (x^3) [/math]
These two functions are different. Which one do you want to solve? The anti-derivative of [imath] \sin (x^3) [/imath] is complicated. Which algorithms do you know to solve it numerically? [imath] (\sin x)^3 [/imath] can be solved by integration by parts.

 

[Dr.Peterson]

Evaluate the definite integral \(\displaystyle \int_{0}^{1} \sin x^3 \ dx\) to the indicated accuracy.
...

if i solve \(\displaystyle \int_{0}^{1} \sin^3 x \ dx\) how to compare it with \(\displaystyle \int_{0}^{1} \sin x^3 \ dx\) when i don't know the result of the original integration?

The question as stated is about \(\displaystyle \int_{0}^{1} \sin(x^3) \ dx\). The question at the end compares two different integrals. Is that the rest of the problem as given to you? If not, why do you ask?

The red is [imath]\sin(x^3)[/imath]; the blue is [imath]\sin^3(x)[/imath]:

​

Since [imath]\sin(x^3)>\sin^3(x)>0[/imath] for [imath]0<x<1[/imath], [imath]\int_{0}^{1} \sin^3 x \ dx<\int_{0}^{1} \sin x^3 \ dx[/imath].

 

[logistic_guy]

[math] (\sin x)^3=\sin^3 x \neq \sin x^3=\sin (x^3) [/math]
These two functions are different. Which one do you want to solve? The anti-derivative of [imath] \sin (x^3) [/imath] is complicated. Which algorithms do you know to solve it numerically? [imath] (\sin x)^3 [/imath] can be solved by integration by parts.

thank

i want to solve \(\displaystyle \sin x^3\)

i don't know algorithms. i just read about simpson rule. it's complicated

The question as stated is about \(\displaystyle \int_{0}^{1} \sin(x^3) \ dx\). The question at the end compares two different integrals. Is that the rest of the problem as given to you? If not, why do you ask?

The red is [imath]\sin(x^3)[/imath]; the blue is [imath]\sin^3(x)[/imath]:

View attachment 38610​

Since [imath]\sin(x^3)>\sin^3(x)>0[/imath] for [imath]0<x<1[/imath], [imath]\int_{0}^{1} \sin^3 x \ dx<\int_{0}^{1} \sin x^3 \ dx[/imath].

thank

the question want me only to solve \(\displaystyle \sin x^3\) to 10 decimal places

i ask because \(\displaystyle \sin x^3\) and \(\displaystyle \sin^3 x\) are the same at the begin and end of the integration limit so i thought they give the same solution but i'm not sure and i want to be sure to compare between them

 

[Dr.Peterson]

the question want me only to solve \(\displaystyle \sin x^3\) to 10 decimal places

i ask because \(\displaystyle \sin x^3\) and \(\displaystyle \sin^3 x\) are the same at the begin and end of the integration limit so i thought they give the same solution but i'm not sure and i want to be sure to compare between them

No, the two functions are not equal at the limits of integration, as is clear in my graph; \(\displaystyle \sin 1^3=\sin 1 = 0.84147...\) and \(\displaystyle \sin^3 1 = (\sin 1)^3 = 0.84147^3=0.59582...\). And that would be irrelevant to accurately working out the definite integral, anyway.

I would expect that whatever source told you to evaluate that integral so accurately has provided some method; I doubt they want you to do it by hand, though. What is the context? And why would you want to do this?

 

[logistic_guy]

No, the two functions are not equal at the limits of integration, as is clear in my graph; \(\displaystyle \sin 1^3=\sin 1 = 0.84147...\) and \(\displaystyle \sin^3 1 = (\sin 1)^3 = 0.84147^3=0.59582...\). And that would be irrelevant to accurately working out the definite integral, anyway.

I would expect that whatever source told you to evaluate that integral so accurately has provided some method; I doubt they want you to do it by hand, though. What is the context? And why would you want to do this?

we're studying complicated functions we can't integrate normally
and we've to figure a way of how to solve them to some given accuracy
they don't give us any method. we're free to use anything but we've to show our steps
i look at Simpson's rule, but i can't use it because it's too complicated
is there another a way to solve this integration to 10 decimal places.
i think even if i'm able to use Simpson rule it can't give this accuracy. am i write?

 

[Dr.Peterson]

we're studying complicated functions we can't integrate normally
and we've to figure a way of how to solve them to some given accuracy
they don't give us any method. we're free to use anything but we've to show our steps
i look at Simpson's rule, but i can't use it because it's too complicated
is there another a way to solve this integration to 10 decimal places.
i think even if i'm able to use Simpson rule it can't give this accuracy. am i write right?

Who is "we" and "they"? I don't think you've mentioned this context before, that I've seen.

If you want accuracy, you need to use something like Simpson's rule, and the more complicated, the more accurate, in general! I'd try it in a spreadsheet or a program.

Why would you think it can't give such accuracy?

 

[logistic_guy]

Who is "we" and "they"? I don't think you've mentioned this context before, that I've seen.

If you want accuracy, you need to use something like Simpson's rule, and the more complicated, the more accurate, in general! I'd try it in a spreadsheet or a program.

Why would you think it can't give such accuracy?

we student
they teacher

what is spreadsheet and what is is program? do they show the steps of the solution?

i think Simpson don't give the 10 accuracy because it has a special formula to calculate the accuracy

\(\displaystyle \frac{(b - a)^5}{180n^4} [\text{max}|f^{(4)}(x)|]\leq \) accuracy \(\displaystyle = \frac{1}{10^{10}}\)

to get 10 decimal places i need to solve \(\displaystyle n\),

the problem \(\displaystyle [\text{max}|f^{(4)}(x)|] = 0\) for the integration in the original question and i can't solve \(\displaystyle n\) so simpson rule isn't let me know the accuracy
this problem let me think simpson rule won't give me the accuracy i want

 

[logistic_guy]

i'll show you what i do

\(\displaystyle f(x) = \sin x^3\)

\(\displaystyle n = 2\)

\(\displaystyle h = \frac{b - a}{n} = \frac{1 - 0}{2} = \frac{1}{2}\)

\(\displaystyle y_0 = f(0) = 0\)
\(\displaystyle y_1 = f(\frac{1}{2}) = 0.1246747334\)
\(\displaystyle y_2 = f(1) = 0.8414709848\)

\(\displaystyle \int_{0}^{1}\sin x^3 \ dx = \frac{h}{3}[(y_0 + y_2) + 4y_1] = \frac{\frac{1}{2}}{3}[(0 + 0.8414709848) + 4(0.1246747334)] = 0.2233616531\)

\(\displaystyle n = 4\)

\(\displaystyle h = \frac{b - a}{n} = \frac{1 - 0}{4} = \frac{1}{4}\)

\(\displaystyle y_0 = f(0) = 0\)
\(\displaystyle y_1 = f(\frac{1}{4}) = 0.0156243642\)
\(\displaystyle y_2 = f(\frac{2}{4}) = 0.1246747334\)
\(\displaystyle y_3 = f(\frac{3}{4}) = 0.4094717771\)
\(\displaystyle y_4 = f(1) = 0.8414709848\)

\(\displaystyle \int_{0}^{1}\sin x^3 \ dx = \frac{h}{3}[(y_0 + y_4) + 4(y_1 + y_3) + 2y_2] = \frac{\frac{1}{4}}{3}[(0 + 0.8414709848) + 4(0.0156243642 + 0.4094717771) + 2(0.1246747334)] = 0.2326004181\)

\(\displaystyle n = 6\)

\(\displaystyle h = \frac{b - a}{n} = \frac{1 - 0}{6} = \frac{1}{6}\)

\(\displaystyle y_0 = f(0) = 0\)
\(\displaystyle y_1 = f(\frac{1}{6}) = 0.0046296131\)
\(\displaystyle y_2 = f(\frac{2}{6}) = 0.0370285701\)
\(\displaystyle y_3 = f(\frac{3}{6}) = 0.1246747334\)
\(\displaystyle y_4 = f(\frac{4}{6}) = 0.2919799046\)
\(\displaystyle y_5 = f(\frac{5}{6}) = 0.5469391732\)
\(\displaystyle y_6 = f(1) = 0.8414709848\)

\(\displaystyle \int_{0}^{1}\sin x^3 \ dx = \frac{h}{3}[(y_0 + y_6) + 4(y_1 + y_3 + y_5) + 2(y_2 + y_4)] = \frac{\frac{1}{6}}{3}[(0 + 0.8414709848) + 4(0.0046296131 + 0.1246747334 + 0.5469391732) + 2(0.0370285701 + 0.2919799046)] = 0.2335812229\)

i don't know when to stop

 

[fresh_42]

I get [imath] 0.2338452455938166 [/imath] from https://www.integral-calculator.com/ with an error of [imath] \pm 2.596203758573797⋅10^{-15} [/imath]

My first thought was to develop the Taylor series at [imath] x=1 [/imath] and work with the series. But I also know an example ([imath] \tan^{-1} [/imath]) where the series expansion is very, very slow and a trig function formula yields a far better convergence.