How do you integrate #int sin^5t/sqrtcost#?

1 Answer
Jan 14, 2017

#intsin^5t/sqrtcostdt=-2(cost)^(1/2)+4/5(cost)^(5/2)-2/9(cost)^(9/2)+C#

Explanation:

Sine and cosine functions can frequently be "turned into" one another, especially with the identity #sin^2t+cos^2t=1#. In this case, we will be using the form #sin^2t=1-cos^2t# to take sine terms and turn them into predominantly cosine forms.

Here, we see that #sin^5t=sin^4tsint=(sin^2t)^2sint#. Using the identity, we see that #sin^5t=(1-cos^2t)^2sint#.

Then,

#intsin^5t/sqrtcostdt=int((1-cos^2t)^2sint)/sqrtcostdt#

The whole point of doing this has been to rewrite the integrand into one trig function completely, except while leaving a single of the other function hanging around. This primes us for the substitution #u=cost#. This implies that #du=-sintdt#.

#=-int((1-cos^2t)^2(-sint))/sqrtcostdt=-int(1-u^2)^2/sqrtudu#

At this point, expand the squared binomial then divide by #sqrtu#.

#=-int(1-2u^2+u^4)/u^(1/2)du=-int(u^(-1/2)-2u^(3/2)+u^(7/2))du#

Integrate term-by-term using #intu^ndu=u^(n+1)/(n+1)#:

#=-(u^(1/2)/(1/2)-2(u^(5/2)/(5/2))+u^(9/2)/(9/2))#

#=-(2u^(1/2)-4/5u^(5/2)+2/9u^(9/2))#

From #u=cost#:

#=-2(cost)^(1/2)+4/5(cost)^(5/2)-2/9(cost)^(9/2)+C#