1

I wasn't able to find this question on another post, and I apologize if its out there.

I have been trying to solve this integral and am not sure how to start. $$\int \sin^3 x dx$$

Thanks

choco_addicted
  • 8,786
  • 8
  • 24
  • 53
Is12Prime
  • 447

5 Answers5

2

Hint: start with the identity $sin^2(x) = 1- cos^2(x)$, break up the integral, and then use a u-substitution with $u=cos(x) $.

Quinn Greicius
  • 1,296
  • 1
  • 9
  • 15
2

recall that $\sin^2 x = 1 - \cos^2x$. And that $\frac{d(\cos x)}{dx} = -\sin x$ or $d(cosx) = -\sin x dx$. \begin{align} \int \sin^3 x dx &= \int (1 - \cos^2 x)\sin x dx\\ &= \int (\cos^2x - 1)(-\sin x dx)\\ &= \int (\cos^2x - 1)d(\cos x)\\ &= \frac{\cos^3 x}{3} - \cos x + c \end{align}

Jeevan Devaranjan
  • 2,575
1

Use the triple angle formula:

$$\boxed{ \phantom\int \sin(3x) = 3\sin(x)-4\sin^3(x)\phantom\int }$$

Surb
  • 55,662
0

$$ \int \sin^3 x dx = \int \sin x \sin^2 x \\ = \int \sin x \left( 1 - \cos^2 x \right) dx \\ = \int \sin x dx - \int \sin x \cos^2 x dx \\ = - \cos x - \int -u^2 dx \\ = - \cos x + \frac{u^3}{3} + C \\ = - \cos x + \frac{\cos^3 x}{3} + C $$

To check if this is right, we can take the derivative.

nukeguy
  • 1,160
  • In general, any time you have an odd power of sine or cosine, factor one out to use with the derivative, use $sin^2(\theta)+ cos^2(\theta)= 1$ to change the remaining even power to the other trig function (sine to cosine and cosine to sine). That is usually one section of the chapter on "integration of trig functions" in any Calculus text! – user247327 Mar 04 '16 at 16:25
0

$$\int \sin^3 x dx$$ $$=\frac{1}{4}\int 4\sin^3 x dx$$ $$=\frac{1}{4}\int (3\sin x-\sin 3x) dx$$ $$=\frac{1}{4} (-3\cos x+\frac{\cos 3x}{3})+c$$

SchrodingersCat
  • 24,472