indefinite integral

indefinite integral

Postby bedje » Mon Jun 01, 2020 1:03 am

Hello. Can you tell me how to solve this integral?
[tex]\int \frac{ (\sin{3x}) ^{4} }{ \sqrt[3]{(\cos{x}) ^{2} } } \cdot dx[/tex]
Posts: 1
Joined: Mon Jun 01, 2020 12:55 am
Reputation: 0

Re: indefinite integral

Postby HallsofIvy » Fri Oct 02, 2020 8:54 pm

My first thought would be to convert that "sin(3x)" to terms of sin(x) only using the identity [tex]sin(3x)=3sin(x)- 4sin^3(x)[/tex] so that integral becomes [tex]\int \frac{3sin(x)- 4sin^3(x)}{cos^{2/3}(x)}dx[/tex]

Note that sin(x) is to odd powers! The standard method is to factor out one "sin(x)" to use with the "dx", then use "[tex]sin^2(x)= 1- cos^2(x)[/tex] to convert everything to cos(x).

[tex]3\int \frac{sin(x)}{cos^{2/3}}dx[/tex]. Let u= cos(x) so that du= -sin(x)dx and the integral becomes [tex]-3\int u^{-2/3}du= -3(-3/2) u^{1/3}+ C= \frac{9}{2}cos^{1/3}(x)+ C[/tex].

And for [tex]-4\int \frac{sin^2(x)}{cos^{2/3}(x)}(sin(x) dx)= -4\int\frac{1- cos^2(x)}{cos^{2/3}(x)}(sin(x) dx)[/tex] let u= cos(x) so the integral becomes [tex]4\int (1- u^2)u^{-2/3}du[/tex][tex]= 4\int u^{-2/3}- u^{4/3} du[/tex][tex]= 4(3u^{1/3}- (3/7)u^{7/3})+ C[/tex][tex]= 12 cos^{1/3}(x)- \frac{12}{7} cos^{7/3}(x)+ C[/tex].

Posts: 341
Joined: Sat Mar 02, 2019 9:45 am
Reputation: 120

Return to Calculus - integrals, lim, functions

Who is online

Users browsing this forum: No registered users and 1 guest