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Current time:0:00Total duration:5:47

Video transcript

so let's see if we can evaluate the indefinite integral x over two times sine of two x squared plus two DX and try to pause the video and see if you can work through it on your own so let's see what is going on here so I have this x over 2 and then I have sine of two x squared plus two now if I were just taking if I were just taking the indefinite integral of sine of X that is that is pretty straightforward the indefinite integral of sine of X well we know that the derivative we know the derivative of cosine of X is equal to negative sine of X so if I were to take the derivative of negative cosine of X that's going to be positive sine of X so this is going to be negative cosine of X and I could have made that even clearer I could have put a negative here and then a negative here and you say okay well look the antiderivative of negative sine of X is just cosine of X and then I have this negative out here negative cosine of X but that's not what I have here I don't have sine of X I have sine of two x squared plus two but then I have this other thing with an X here and so what your brain might be doing or it's good once you get enough practice when your brain will start to do this say okay well this is interesting this kind of looks like the derivative of this if this is if we were to call this and I'm tired of that orange if we were to if we were to call a why this color changing is if we were to call this f of X if two x squared plus two is f of X - x squared plus two is f of X what is f prime of X well then F prime of X F prime of X is going to be four X and this thing right over here isn't exactly 4x but we can make it we can we can do a little rearranging multiplying and dividing by a constant so this becomes 4x what if what if we were to what if we were to multiply and divide by four so we multiply by four there and then we divide by four and then we take it out of the integral sign and even better let's take this two out so let's just take so this is let's take the one-half out of here so this is going to be one-half and so I could have rewritten the original integral as 1/2 times 1/4 so it's 1/8 times the integral times the integral of 4x times sine of 2x squared plus 2 DX well now this is interesting because if this is f of X if if this business right over here is f of X so we're essentially taking sine of f of X then this business right over here is f prime of X which is a good signal to us at a the reverse chain rule is applicable over here we can we can rewrite this we could also rewrite this is this is going to be equal to 1 we can rewrite this as the same thing as 1/8 I went to the same colors I'm using a new art program and sometimes the color changing isn't as obvious as it should be so 1/8 times the integral of f prime of X F F prime of x times sine sine of f of X sine of f of X DX let me throw that f of X in there so sine of f of X and so when you view it this way you say hey you know by the reverse chain rule I have IIIi have a function and I have its derivative here so I can really just take the antiderivative with respect to this this is essentially what we're doing a new substitution you could do use substitution here you could set u equaling this and then D U is going to be 4x DX but now we're getting a little practice starting to do it a little bit more in our heads so what would this integral evaluate out to be well this would be this would be 1/8 x times well if you take the antiderivative of sine of f of X with respect to f of X well we already saw that that's negative cosine of X so this is going to be times negative cosine negative cosine of f of X negative cosine of f of X negative cosine of f of X whoops put a blue there and then of course you have your plus C so what is what is this going to be well instead of just saying F Prime instead of saying it in terms of f of X we just say it in terms of 2x squared this is going to be or 2x squared plus 2 it's going to be one eighth times I have a function and I have its derivative so I can really just integrate with respect to that function so it's going to be times negative negative cosine of 2x squared plus 2 - x squared plus two and then of course I have my Y keep switching to that color I have my plus C and of course I could I could just take the negative out it would be negative 1/8 cosine of this business and then plus C and we're done we have just employed the reverse chain rule we could have used we could have used u substitution but hopefully we're getting a little bit of practice here hey I'm seeing something here and I'm seeing its derivative so let me just integrate with respect to this thing which is really what you would set u to be equal to you're integrating with respect to the U and you have your D you hear this this times this is D u so you're saying it's like integrating sine of u D you I encourage you try to use u substitution here and you'll see it's the exact same thing that we just did