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0:00
in this video we're going to focus on trigonometric integrals but now to begin you may want to write down some notes just to review some formulas that will be useful when integrating trigonometric expressions first sine squared plus cosine squared we need to know that equals one next one plus tangent squared is equal to secant squared this is a good review of some trigonometric formulas that you've been exposed to in the past the next one you need to be familiar
0:30
with is one plus cotangent squared is equal to cosecant squared
0:40
next we have the double angle formulas sine two x is equal to two sine x cosine x and then cosine two x there's three forms that you can convert this into you can convert it to cosine squared minus sine squared you could set it equal to one minus two sine squared or two cosine squared minus one next up are the power reducing formulas sine squared can be reduced to one half one minus cosine two x and cosine squared you can reduce that to one half one plus cosine 2x
1:37
now there are some other trigonometric formulas that you may need to know which i'll introduce later in this video when it becomes relevant but for now we can work with what we have here now the first problem that we're going to work on is this one let's find the anti-derivative of cosine to the third x so what do you think we need to do here what we could do is we can write cosine cube as cosine squared times cosine x and what we want to do is we want to change cosine squared into one minus
2:17
sine squared keep in mind sine squared plus cosine squared is equal to one solving for cosine we can move sine squared to the other side if we do that we'll get that cosine squared is one minus sine squared so we're going to replace cosine squared with one minus sine squared
2:45
now the reason why we want to put it in this form is so that we can do u substitution if we make u equal to sine x d u is going to be cosine x and dx so that's why we have to keep at least one cosine on the outside so that we can use it to change it into d u so this part here is d u and we're going to replace sine with u so this becomes the integral of one minus u squared d u the antiderivative of one is going to be u the antiderivative of u squared is u to the third over three and then plus c
3:33
now our last step is to replace u with sine so our final answer is going to be sine x minus 1 over 3 sine to the third x plus c so that's the antiderivative of cosine cubed
3:53
now let's work on another similar problem but one that's a little bit harder cosine raised to the fifth of x go ahead and find the anti-derivative of that expression so what we're going to do is we're going to take out a cosine so this is going to be cosine to the fourth times cosine x dx now we want this to become eventually du like we did before so we need to change cosine into sine but we need to get cosine squared first cosine to the fourth is cosine squared squared
4:37
now cosine squared we can make that one minus sine squared
4:52
now at this point we can replace sine with u
5:00
d u is going to be cosine x and dx so we're going to replace this with d u so this becomes the integral of one minus u squared raised to the second power and all of this can be replaced with d u
5:20
now before we can integrate this expression we need to expand it so what we have is one minus u squared times one minus u squared so let's foil one times one and that's going to be 1 and then this is negative u squared plus another negative u squared so that's going to be negative 2 u squared and then negative u squared times negative u squared that's gonna be positive u to the fourth and then times du
6:07
so now let's go ahead and find the antiderivative of this expression so it's going to be u minus two u to the third over three plus u to the fifth power over five using the power rule and then plus the constant of integration so now we can write our final answer let's replace u with sine so we're gonna have sine x minus two over three sine to the third power of x and then plus one over five sine to the fifth power of x plus c
6:47
so this is the indefinite integral of cosine x raised to the fifth power so this is our final answer
6:58
now let's try another problem let's find the indefinite integral of cosine to the fifth x times sine x dx feel free to try this problem
7:13
so what should we do here notice that we have a sine by itself so what we want to do is we want to make u equal to cosine because d u is going to have the sign that we need to cancel this sign the derivative of cosine is negative sine and if we solve for dx it's going to be d u divided by negative sine so we can go ahead and replace cosine with u so this would be u to the fifth power and then we could replace dx with what we have here d u over negative sine x sine will cancel
7:56
so we're going to have the integral of negative u to the fifth d u and then using the power rule it's going to be negative u to the 6 over 6 and then plus c so now we just got to replace u with cosine and so our final answer is going to be negative one over six cosine to the sixth power of x plus c so this is the answer this problem wasn't too bad it was pretty straightforward now let's try a similar problem that is going to be a little bit longer than the other one
8:34
so let's find the indefinite integral of sine to the fifth power times cosine squared of x dx feel free to pause the video if you want to now the question is should we make u equal to sine or cosine cosine has an even power sine has an odd power what we want to do is we want to make u equal to cosine because what we could do is take out a sine leave it by itself and then the sine to the fourth we can convert all of that into cosine so let me show you sine to the fifth we're going to write
9:12
that as sine to the fourth x times sine times cosine squared this sign we want that to pair up with the dx because that's going to be our du everything else we want it to be not in terms of sine but cosine sine to the fourth i'm going to write that as sine squared squared and then we have cosine squared and i'm going to put sine next to dx
9:49
now we know that sine squared is one minus cosine squared based on the trig identity sine squared plus cosine squared is one
10:02
and now we have everything in terms of cosine except this particular trig function sine so now we can make u equal to cosine d u is going to be negative sine x dx so dx is this
10:27
so everywhere we see a cosine we're going to replace it with u so this is going to be 1 minus u squared squared times u squared times sine x and then let's replace dx with d u over negative sine x so these will cancel
10:51
and what we need to do next is we need to foil
11:05
so this is what we have at this point and let's not forget the negative sign that we have here so i'm going to put that in the front now we've already foiled 1 minus u squared times 1 minus u squared based on a previous problem we know it's going to be 1 minus 2u squared plus u to the fourth our next step is to distribute u squared so u squared times one that's going to be u squared but let's go ahead and distribute this negative sign as well so it's going to be negative u squared
11:45
and then u squared times negative 2 u squared times the negative sign that's going to be positive 2 u to the fourth and then u squared times u to the fourth with the negative in front that's going to be negative u to the sixth
12:05
so now let's go ahead and find the antiderivative of each of these terms so it's gonna be negative u to the third over three plus two u to the fifth over five minus u to the seventh over seven plus c
12:23
so now let's go ahead and replace u with cosine so it's going to be negative one over three cosine to the third power plus two over five cosine to the fifth power minus one over seven cosine to the seventh power plus c so this right here is going to be our final answer so that is the antiderivative of sine to the fifth x times cosine squared x here's another similar problem let's find the indefinite integral of sine to the fifth power of x times cosine cube x try that one
13:10
now we got to decide if we want to make u equal to sine or cosine because we can isolate a sine from the integral we can split that into sine to the fourth times sine or we could split this into cosine squared times cosine i think it's easier if we split cosine cubed into cosine squared times cosine and then convert the cosine squared into one minus sine squared leaving that last cosine to be with dx let's do it that way because to change this to sine to the fourth times sine and then to convert
13:48
sine to the fourth into sine squared times sine squared that's gonna be more work if we focus on this one if we split it apart and since it's easier to deal with three than five we don't have to foil so it's easier if we break up cosine cubed into cosine squared and cosine
14:14
now all we need to do is change cosine squared into one minus sine squared
14:25
so now we have everything in terms of sine except one cosine function so now let's go ahead and make u equal to sine x and d u will simply be cosine x dx so this is going to be u to the fifth
14:50
and then times 1 minus u squared and then cosine x dx we could simply replace that with du now let's distribute the u to the fifth to one minus two squared so it's going to be u to the five minus u to the seven and then d u the anti-derivative of u to the five is going to be u to the six over six and then for u to the seventh that becomes u to the eight over eight plus c now let's replace u with sine x so the final answer is going to be one over six sine to the sixth power of x
15:32
minus one over eight sine to the eighth x plus c so this is the indefinite integral of the original problem now the next problem that we need to work on is finding the indefinite integral of sine squared x
15:57
now this problem is definitely different from what we've been doing before so what do you recommend that we should do here because it's not really advantageous to convert this into one minus cosine squared in this case we need to use the power reducing formulas if you recall earlier at the beginning of the video we wrote that sine squared is one half times one minus cosine two x so what we could do is move the one half to the front
16:41
and focus on what we have here the anti-derivative of one is going to be let's put this in parentheses x now the antiderivative of cosine 2x the antiderivative of cosine is sine but for negative cosine it's going to be negative sine 2x but we need to divide it by the derivative of the inside the derivative of 2x is 2. so we've got to divide this by 2. so that's the antiderivative of negative cosine 2x it's negative sine 2x divided by 2.
17:22
now all we need to do at this point is distribute the one-half we could leave our answer like this if we want to but let's go ahead and distribute the one half so it's going to be one half x minus 1 4 sine 2x plus c so this is the antiderivative of sine squared using a double angle now keep in mind we can always replace sine 2x with 2 sine x cosine x using the double angle formula but let's go ahead and move on to the next problem let's find the antiderivative of
18:02
instead of cosine squared let's make this cosine squared of 3x so go ahead and try that so first we need to talk about the power reducing formula of cosine squared cosine squared x is one half one plus cosine two x so notice the angle it went from x to 2x it doubled what we have here is cosine squared 3x so using the power reducing formula the angle is going to double it's going to go from 3x to 6x so this is going to be one half times 1 plus cosine 6x so we're going to have one half integral
18:55
1 plus cosine 6x and then dx so now we can find the antiderivative the antiderivative of 1 is going to be x and the anti-derivative of cosine 6x that's going to be sine 6x divided by the derivative of 6x which is 6 and of course plus c so now the last thing that we need to do is distribute the one half so we're gonna have one half x plus one over twelve sine six x plus c
19:40
this right here is the final answer
19:45
now for the next problem we're going to try finding the indefinite integral of sine to the fourth x dx so go ahead and try that problem
19:57
so this is the integral of sine squared x and that is going to be squared as well sine squared raised to the second power is the same as sine to the fourth and now we could use the power reducing formula on sine squared
20:18
so it's going to be a half times 1 and then not plus but minus cosine 2x
20:30
and then all raised to the second power now we're going to distribute the exponent so if we distribute to one-half it's going to be one-half squared so that becomes one-fourth
20:46
and then we're gonna have 1 minus cosine 2x raised to the second power which means we need to foil that now we can move the 1 4 to the front so we have 1 over 4 integral 1 minus cosine 2x times 1 minus cosine 2x dx
21:23
so first we have one times one which is one and then this is going to be negative cosine 2x plus another negative cosine 2x that's negative 2 cosine 2x and then negative cosine 2x times negative cosine 2x that's positive cosine squared of 2x
21:49
now before we integrate this expression let's use the power reducing formula on cosine square 2x so we know cosine squared of 1x is one-half 1 plus cosine 2x so therefore cosine squared of 2x is going to be one half one plus cosine four x so the key here is to realize that when you're using the power of reducing formula the angle doubles so if it doubles from 1x to 2x then if we start from 2x it's going to double to 4x
22:35
so this becomes 1 4 integral
22:41
one minus two cosine two x and then plus one half well let's distribute the half so this is one half times one or just one half and then this is going to be plus one half cosine four x
23:09
now what we could do is combine like terms one plus one half one is two over two so plus one over two that becomes three over two
23:25
so we can just put 3 over 2 here and now let's find the anti-derivative so the anti-derivative of negative 2 cosine 2x that's going to be negative 2. the antiderivative of cosine is negative sine i mean not negative sine but positive sine 2x divided by 2. the antiderivative of 3 over 2 is just 3 over 2 times x and for cosine to the 4x is going to be sine to the 4x over 4. and then of course plus c now let's organize our answer so i'm going to keep the one-fourth in the front
24:10
i'm going to write this one first so plus or just three over two x now these twos will cancel so we're just going to have minus sine 2x and this is going to be 1 over 8 sine 4x
24:36
and then plus the constant of integration so this is the antiderivative of sine to the fourth x now let's try this problem what is the anti-derivative of sine squared cosine squared dx go ahead and try that so for this problem we need to use the power reducing formulas on sine squared and cosine squared so sine squared is one half times one minus cosine two x cosine squared is one half but one plus cosine two x one half times one half is one over four so we can move that to the front
25:29
and then we need to foil so here we have one times one so that's going to be one and then positive cosine two x will cancel with negative cosine two x so the two middle terms cancel and then negative cosine times positive cosine is going to give us negative cosine squared of 2x
25:54
1 minus cosine squared that's an identity one minus cosine squared is equal to sine squared so we now have is one fourth integral sine squared 2x dx now if sine squared 1x is this sine squared 2x must be the same thing but with a 4x here instead of a 2x so we now have is 1 4 integral and then using the power reducing formula this is going to be one half times 1 minus cosine but not 2x this is going to double the 4x
26:41
now one-fourth and one-half we can combine so it's one over eight integral
26:51
1 minus cosine 4x dx
26:57
the antiderivative of 1 is x the antiderivative of cosine is going to be sine but 4x divided by the derivative of 4x which is 4. and then plus c so this is the final answer to the problem for the next problem we're going to find the indefinite integral of cosine squared times tangent to the third power so what do you think we need to do here in order to find the indefinite integral of this expression well we don't know of an identity that involves cosine and tangent so the best
27:48
thing to do is to convert tangent into sine over cosine
27:57
so tangent cube is going to be sine to the third divided by cosine cubed or we could write cosine cube as cosine squared times cosine
28:14
so that way we can easily cancel out a cosine squared
28:21
now if we try to make u equal to sine d u is going to be cosine dx but the cosine is on the bottom not on the top so it's not going to cancel well the way it is so we want to make u equal to sine i mean u equal to cosine that sine so we're going to do is we're going to split up sine cube into sine squared times sine
28:53
and we'll leave cosine on the bottom for now now in the next step we're going to change sine squared into 1 minus cosine squared
29:16
now at this point we can make u equal to cosine x d u is going to be negative sine x dx and so dx is going to be d u over negative sine x so this becomes 1 minus u squared divided by u and then we have sine x times d u over negative sine
29:52
so we can cross that out and then let's move the negative sign to the front so we have negative integral let's break this up into two smaller fractions so it's going to be 1 over u
30:11
and then minus u squared over u which is negative u
30:20
so this is going to equal negative lnu and then these two negative signs they'll cancel so that's going to be positive and the anti-derivative of u is going to be u squared over two plus c
30:43
so this is going to be negative ln and then uh absolute value cosine x plus one half cosine squared x plus c so this is the final answer to uh the problem at hand that's it that's all we got to do here
31:27
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