How do you integrate #int sin^5t/sqrtcost#?
1 Answer
Explanation:
Sine and cosine functions can frequently be "turned into" one another, especially with the identity
Here, we see that
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
#=-int((1-cos^2t)^2(-sint))/sqrtcostdt=-int(1-u^2)^2/sqrtudu#
At this point, expand the squared binomial then divide by
#=-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
#=-(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
#=-2(cost)^(1/2)+4/5(cost)^(5/2)-2/9(cost)^(9/2)+C#