![]() Plus C's and all of that but remember, this is gonnaīe a definite integral so all of those arbitraryĬonstants are going to get canceled out. So g prime of x is equal to cosine of x in which case g of x, theĪnti-derivative of cosine of x, well it's just sine of x orĪnother way to think about it, the derivative of sine The anti-derivative ofĬosine of x is sine of x. Is going to be equal to one, and then what would my g prime of x be? Well my g prime of x,Ĭosine of x if I take its anti-derivative, itĭoesn't get more complicated. Say is x, in which case, f prime of x, f prime of x, Write that over here, so my f of x, I will Just one, so I'm gonna make that my f of x, so I could I take its derivative? Well the derivative of x is Between x and cosine of x, which one gets more simple when Maybe be easier to find the anti-derivative of. I take its anti-derivative, then this expression will G prime of x does not get more complicated when Simplified when I take its derivative, and the ![]() I take its anti-derivative, it doesn't get more complicated. Prime of x that when I take its anti-derivative, so when When I take its derivative, it simplifies it, so simplify, and you wanna find a g And just to reiterate what I said before, you wanna find an f of x that You then swap these around, minus f prime of x, g of x, dx. This is gonna be equal to f of x times g of x minus, Just comes straight out of the product rule that you learned in differential calculus, To, and in other videos we prove this, it really So if I have f of x times g prime of x, dx, this is going to be equal I'll do it right over here, if I have the integralĪnd I'll just write this as an indefinite integralīut here we wanna take the indefinite integralĪnd then evaluate it at pi and evaluate it at zero, ![]() So let's just remind ourselvesĪbout integration by parts. That we should be using integration by parts. Like x if you were to take its derivative, it gets simpler. ![]() Take the anti-derivative of without making it moreĬomplicated like cosine of x, and another of the functions One of these functions is fairly straightforward to See a product of functions right over here, and if We're probably going to have to use a slightly more Just straight up take the anti-derivative hereĪnd then evaluate that at pi and then subtract from that and evaluate it at zero, so Well when you immediately look at this, it's not obvious how you Like always, pause this video and see if you can evaluate it yourself. Gonna do in this video is try to evaluate theĭefinite integral from zero to pi of x cosine of x dx. ![]()
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