Integral of 1/(2+tanx)

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Trig integrals always come with sneaky tricks. This one is no exception!

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I too took the route to yield arcsec (3x) + c and love how you search for alternative pathways to the solution. Mathematics is all about blazing a trail and making it to the destination through any consistent model. Ultimately, mathematics is about innovation and doing what you have never done before by being resourceful and not giving up.

mathematix-rodcast

The way I did this was to split the numerator into A(sin x + 2 cos x) + B(cos x - 2 sin x), where the 2nd term is the dertivative of the denominator.

dyld

I found another way to solve this problem--also a pretty cumbersome technique but it works out nicely! Substitute u=2+tanx, du=sec^2xdx, dx=cos^2xdu. Form a triangle to find that cos^2x = 1/(u-2)^2+1. Put everything back into the integral, change the bounds. You get integral from 2 to 3 of 1/u((u-2)^2+1). Use partial fractions to separate. The first part is 1/5(ln(3/2). The second part is trickier. The integral is -1/5(u-4/u^2-4u+5). Can't substitute because if you derive the denominator you would get 2u-4. Noticing that 2u-4 is just twice u-2, I asked myself, wouldn't it be nice if the numerator was just u-2?. Then I used the linearity of the integral to split it up into the integral of -1/5(u-2/u^2-4u+5) and -1/5(-2/u^2-4u+5). You can use arctangent antiderivative for the second part, and do a basic substitution for the first part. Evaluating everything and doing some manipulation, you'll end up with the same result! I liked this method because it involves partial fractions, two substitutions, and the technique bprp likes to call "wouldn't it be nice?"

AKSatMusic

We can take the substitution y=tgx, then1/(cosx)^2=1+y^2, then we have integral from (1/(2+y))*(1/1+y^2) dy where y go from0 till pí/4.Then[ 1/(2+y)] *[1/(1+y^2)]=A/(2+y) + (By+C)/(1+y^2), where will get the same result, ther will be functions ln

tgx

or use the lamda, mu and r relation. (denominator)'+r and find the three constants then it will be simplified and use u'/u to get lnu .

tangpiseth

I thought you were going to use the substitution u = pi/2 - x or something like that, the "I = " method that bprp, flammy, and others have been posting a lot about. So, this was a nice unexpected thing that I learned from!

txikitofandango

I thought you were going to use complex analysis for a minute.

thomasblackwell

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