Evaluate the integral: ∫sin⁡(x3)dx

Question

Evaluate the integral:  ∫sin⁡(x3)dx

ol3i4c5s4hr · Accepted Answer

By combining Euler&#039;s formula with the integral expression for the Γ function, we have, for n&amp;gt;1∫0∞sin⁡(xn)dx=(1n)!sin⁡π2nand∫0∞cos⁡(xn)dx=(1n)!cos⁡π2nGiven the fact that your integral is indefinite, its expression is very similar, but it involves the incomplete Γ function, rather than the classical one.

Charles Benedict · Accepted Answer

First note that  ∫sin⁡(x3)dx=12(∫ieix3dx−∫ie−ix3dx)  For the first integral  I1=∫ieix3dx  change variables with u=ix3, x=−iu3 and  dx=−13iduu23  then  I1=−13∫u−23eudu  Now with the incomplete gamma function  γ(s,z)=∫0zts−1e−tdt  we can write I1 as  I1=−13(−u)23u23γ(13,−u)  and substituting back u=ix3  I1=(−ix3)23u23γ(13,ix3)  Similarly for the second integral  I2=∫ie−ix3dx  we have  I2=−(ix3)233x2γ(13,ix3)  Finally  ∫sin⁡(x3)dx  =(−ix3)23γ(13,−ix3)+(ix3)23γ(13,ix3)6x2+C

star233 · Accepted Answer

How to integrate: ∫sin⁡(x3)dx Start with a series representation for: f(x)=sin⁡(x) The taylor&#039;s series representation of sin⁡(x): f(x)=sin⁡(x)=∑n=0∞(−1)n(2n+1)!x2n+1 Substitute x→x3 f(x3)=sin⁡(x3)=∑n=0∞(−1)n(2n+1)!(x3)2n+1 Integrate: I=∫f(x3)dx=∫sin⁡(x3)dx=∫(∑n=0∞(−1)n(2n+1)!(x3)2n+1)dx) I=∑n=0∞((−1)n(2n+1)!(∫x6n+3dx)) I=∑n=0∞(−1)nx6n+4(2n+1)!(6n+4)+C

Evaluate the integral: \int\sin(x^3)dx

Answered question

Answer & Explanation

New Questions in Calculus 1