# indefinite integral

### indefinite integral

Hello. Can you tell me how to solve this integral?
$$\int \frac{ (\sin{3x}) ^{4} }{ \sqrt[3]{(\cos{x}) ^{2} } } \cdot dx$$
bedje

Posts: 1
Joined: Mon Jun 01, 2020 12:55 am
Reputation: 0

### Re: indefinite integral

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

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 "$$sin^2(x)= 1- cos^2(x)$$ to convert everything to cos(x).

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

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

HallsofIvy

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