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## Integrating using trigonometric identities

Current time:0:00Total duration:4:33

# Integral of sin^2(x) cos^3(x)

## Video transcript

- [Voiceover] Let's see if
we can take the indefinite integral of sine squared x
cosine to the third x dx. Like always, pause the
video and see if you can work it through on your own. All right, so right when
you look at it you're like, "Oh wow, if this was just a sine of x, "not a sine squared of x. "Well that's going to be the negative of "the derivative of cosine of x. "Maybe I could have used u-substitution. "Likewise, if this was just
a cosine of x not a cosine "to the third of x, I could
have used u-substitution. "I could have said u
is equal to sine of x, "but I can't do that over here." So the general ideas here,
if one of these has an odd exponent and you see
that this one does have an odd exponent, the
cosine has an odd exponent. What you do is you try to algebraically engineer
this expression inside, so that you can use u-substitution. The way that you can do
that, if you have an odd exponent like this, is to
separate out one of the cosine's x and then use
the Pythagorean Identity with the remaining cosine squared
x, what do I mean by that? Let me just rewrite this,
so this can be rewritten as sine squared x times... Let me write it this way. Times cosine squared x cosine x. All I did is I rewrote
sometime to the third power, something to the second
power times that thing to the first power, dx. And this could be rewritten as, sine squared x, then I'm going to use
the Pythagorean Identity to convert this into
one minus sine squared. So that, by the Pythagorean
Identity, is the same thing as one minus sine squared x and then we have cosine x, times cosine x dx. Now I can distribute this
sine squared times one minus sine squared and I am left with... Let me do this in a new color,
this business right here. Now I'm left with the
indefinite integral of, Sine squared x times one is going to be sine squared x and then sine squared x times negative sine squared x is negative sine to the fourth. Then all of that times cosine x. All of that times cosine x dx. Now this is starting to look interesting, cause I have sine squared x
minus sine to the fourth x, but I have the derivative
of sine sitting out here. I have cosine x, that's
the whole reason why we did this little algebraic manipulation. So u-substitution works
out quite well now. Because if we said that.. Let me do another color, I'll do purple. If we say that u is equal to sine of x, then du is going to be equal to cosine of x dx, and that works out quite well because we have the du right over here. Then this would be u squared
minus u to the fourth. Which we know how to take
the anti-derivative of, it's over the whole stretch. We can rewrite this as
the indefinite integral. Instead of sine squared
x we're saying sine is the same thing as u so
we can rewrite that as, u squared minus u to the fourth times du. This is pretty straight
forward, now this is going to be u to the third over three minus u to the fifth over five plus c. Then we just do the reverse substitution and then that gets us to be, instead of u we want to put a sine of x there. So we're gonna get sine to the third, sine of x to the third over three minus sine of x to the fifth power. Let me rewrite that five a little bit. Sine of x to the fifth power over five plus c and we are done.