How Can I Solve the Integral of Sin^8x Using Integration by Parts?

[iceman]

Hi can anyone help me solve this integral, I'm having trouble with this one?

the integral is: int(sin^{8}x.dx)->upper limit=pi ->lower limit=0.

Q) Evaluate the integral exactly using integration by parts to get a reduction formulae for int(sin^{n}x.dx)

 

[KLscilevothma]

Hint(to find the reduction formula):
[inte]₀^pi sinⁿx dx
= - [inte]₀^pi sin^n-1x d (cos x)
= [- cos x sin^n-1x]₀^pi + [inte]₀^pi cos x d(sin^n-1x)
= [inte]₀^pi cos x d(sin^n-1x)
= ...

 

[tacman]

Sure, I can help you with solving this integral using integration by parts. The first step in solving this integral is to identify which part of the function should be considered as the "u" and which part should be considered as the "dv". In this case, we can let "u" be sin^7x and "dv" be sinx.

Using the integration by parts formula: int(u.dv) = u.v - int(v.du), we can rewrite the integral as:

int(sin^7x.sinx.dx) = sin^7x*(-cosx) - int(-cosx*7sin^6x.cosx.dx)

= -sin^7x*cosx + 7int(sin^6x.cos^2x.dx)

Next, we need to use the trigonometric identity cos^2x = 1/2(1+cos2x) to simplify the integral further.

int(sin^8x.dx) = -sin^7x*cosx + 7int(sin^6x*(1+cos2x).dx)

= -sin^7x*cosx + 7int(sin^6x.dx) + 7int(sin^6x.cos2x.dx)

= -sin^7x*cosx + 7int(sin^6x.dx) + 3.5int(sin^6x.cos2x.dx)

We can now use the reduction formula for int(sin^nxdx) = -1/n*sin^(n-1)x*cosx + (n-1)/n*int(sin^(n-2)x.dx), where n is any positive integer.

Therefore, we can continue to simplify the integral as:

int(sin^8x.dx) = -sin^7x*cosx + 7*(-1/7*sin^6x*cosx + (6/7)*int(sin^4x.dx)) + 3.5*(-1/7*sin^6x*cos2x + (6/7)*int(sin^4x.cos2x.dx))

= -sin^7x*cosx - 1*sin^6x*cosx + (6/7)*int(sin^4x.dx) - 0.5*sin^6x*cos2x + (3/7)*int(sin^4x.cos2x.dx)

= -sin^7x*cosx - sin^6x*cosx + (6/7)*int