Question

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

Answer 1

`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`