What is the integral of #int sin^5 (x) dx#?

Question

Paisley Johnson · Accepted Answer

The answer is #=-1/5cos^5x+2/3cos^3x-cosx+C#

We need

#sin^2x+cos^2x=1#

The integral is

#intsin^5dx=int(1-cos^2x)^2sinxdx#

Perform the substitution

#u=cosx#, #=>#, #du=-sinxdx#

Therefore,

#intsin^5dx=-int(1-u^2)^2du#

#=-int(1-2u^2+u^4)du#

#=-intu^4du+2intu^2du-intdu#

#=-u^5/5+2u^3/3-u#

#=-1/5cos^5x+2/3cos^3x-cosx+C#

Ezra Adams · Answer

undefined To find the integral of $ \int \sin^5(x) \, dx $, you can use the reduction formula for powers of sine functions:

$$ \int \sin^n(x) \, dx = -\frac{\sin^{n-1}(x) \cos(x)}{n} + \frac{n-1}{n} \int \sin^{n-2}(x) \, dx $$

Using this formula repeatedly, you can reduce the power of $ \sin(x) $ until it reaches a known integral. However, this process can become quite complex and iterative for higher powers of $ \sin(x) $.

Another approach is to use trigonometric identities to rewrite $ \sin^5(x) $ in terms of lower powers of $ \sin(x) $ and $ \cos(x) $, and then integrate each term separately. This method can be more straightforward than using the reduction formula directly.

Alternatively, you can also use numerical integration methods if an exact solution is not necessary or feasible.