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let's say we have the indefinite integral of 1 over 36 plus 36 plus x squared DX now as you can imagine this is not an easy integral to solve without trigonometry there's I can't do u substitution I don't have the derivative of this thing sitting someplace this would be easy if had a 2x sitting up there then I would say oh the derivative of this is 2x I could do u substitution and I'd be set but there is no 2x there so how do I do it well I resort to our trigonometric identities and let's see what trig identity we can get here so the first thing I always do and this is just the way my brain works I always like it you know I can see this is a constant plus something squared which tells me I should use a trigonometric identity but I always like it in terms of 1 plus something squared so I'm just going to rewrite my integral as being equal to let me write the DX in the numerator right this is just x DX we write a nicer integral than that so this is equal to the integral of DX over over 36 times 1 plus x squared over 36 1 plus x squared over 36 that's another way to write my integral now let's let's see if any of our trig identities can somehow be substituted in here for that that would somehow simplify the problem so the one that springs to mind and if you don't know this already I'll write it down right here is 1 plus tangent squared of theta 1 plus tangent squared of theta and let's prove this one tangent square root of theta this is equal to 1 plus just the definition of tangent sine squared of theta over cosine squared of theta now 1 is just cosine squared over cosine squared so I can rewrite this as equal to cosine squared of theta over cosine squared of theta that's 1 plus sine squared theta over cosine squared of theta we have a common denominator now what's cosine squared plus sine squared definition of the unit circle that equals 1 over cosine squared of theta or we could say that that equals 1 over cosine squared 1 over cosine is secant so this is equal to the secant squared of theta so if we make the substitution if we say let's make this thing right here let's make that thing equal to tangent of theta or tangent squared of theta then this expression will be 1 plus tangent squared of theta which is equal to secant squared and maybe that'll help simplify this equation a bit so we're going to say that x squared over 36 is equal to tangent squared of theta let's take the square root of both sides of this equation and you get x over 6 is equal to the tangent of theta or that X is equal to 6 tangent of theta if we take the derivative of both sides and this with respect to theta we get DX D theta is equal to what's the derivative of the tangent of theta I can show it to you just by going from these basic principles right here actually let me let me do it for you just just in case so the derivative of tangent theta never hurts to do it on the side let me do it right here so it's going to be 6 times the derivative with respect to theta of tangent of theta which we need to figure out so let's figure it out the derivative of theta well tangent of theta well that's the same thing as DD theta of sine of theta over cosine of theta that's just the derivative of tangent or this is just the same thing as the derivative with respect to theta let me scroll to the right a little bit of because I never remember the quotient rule I've told you in the past that it's somewhat lame of sine of theta times cosine of theta to the -1 power and what is this equal to we could say it's equal to well the derivative of the first expression or the first function we could say which is just cosine of theta so this is equal to cosine of theta that's just the derivative sine of theta times our second expression times cosine of theta to the minus one I've made I put these parentheses and put the minus one out there because I didn't want to put the minus one here make you think that I'm talking about an inverse cosine or an arc cosine so that's the derivative of sine times cosine and now I want to take plus the derivative of cosine so what's the derivative of well not just cosine the derivative cosine of the minus 1 so that is minus 1 times cosine to the minus 2 power of theta that's the derivative of this of the outside times the derivative of the inside so let me scroll over more so that's the derivative of the outside if the cosine theta was just an X you would say X to the minus ones derivative is minus 1 X to the minus 2 now times the derivative of the inside of cosine of theta with respect to theta so that's x minus sine of theta and I'm going to multiply all of that times sine of theta times sine of theta the derivative of this thing which is stuff in green times the first expression so what is this equal these cosine of theta divided by cosine theta that is equal to 1 and then I have a minus 1 and then I have a minus sine of theta so that's plus plus and what do I have sine squared sine of theta times sine of theta over cosine squared so plus sine squared of theta over cosine squared of theta which is equal to 1 plus tangent squared of theta well what's 1 plus tangent squared of theta I just showed you that that's equal to secant squared of theta so the derivative of tangent of theta is equal to secant squared of theta all that work to get us fairly something well it's nice when it comes out simple so DX D theta this is just equal to secant squared of theta so if we want to figure out what DX is equal to DX is equal to just both sides times D theta so it's 6 times secant squared d theta-- so that's our DX and of course in the future we're going to have to we're going to back substitute so we want to solve for theta so that's fairly straightforward you just take the arctangent on both sides of this equation you get the arc tangent of x over 6 is equal to theta and we'll save this for later so what is our integral reduced to our integral now becomes the integral of DX what's DX it is 6 secant squared theta D theta all of that over this this denominator which is 36 times 1 plus tangent squared of 1 plus tangent squared of theta now we know that this right there is secant squared of theta I've shown you that multiple times so this is secant squared of theta in the denominator we have a secant squared of theta in the numerator they cancel out so those cancel out and so our integral reduces to lucky for us 6 over 36 which is just 1 over 6 D theta which is equal to 1 over 6 theta plus C and now we back substitute using this result theta is equal to arctangent of X over 6 so the antiderivative of 1 over 36 plus x squared is equal to 1/6 times theta theta is just equal to the arc tangent the arc tangent of x over 6 plus C and we're done so that one wasn't too bad