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 Skip Navigation LinksMath Help > Calculus > Integral > Integral of sqrt(tan(x))

Proofs of Selected Integrals

òsqrt(tan(x)) dx
Solution, with a big hint by David Loeffler:

I will find sqrt(tan(x)) using the substitution v = sqrt(tan(x))

First, a little differentiation of v:

v = sqrt(tan(x))
dv = sec2(x) / (2 sqrt(tan(x)))
dv = 1 / (2 cos2(x) sqrt(tan(x)))

Now, I will process the integral so that dv appears in it:

sqrt(tan(x)) dx
sqrt(tan(x)) dx/ (cos2(x) + sin2(x))
sqrt(tan(x)) dx/ (cos2(x) (1+tan2(x)))
tan(x) dx/ (cos2(x) sqrt(tan(x)) (1+tan2(x)))
2tan(x) dx/ (2 cos2(x) sqrt(tan(x)) (1+tan2(x)))
2v2 dv / (1+v4)

David took us this far, and then gave one more hint, which led us to express 1+v4 this way:

1+v4 = (v2-sqrt(2)v+1)(v2+sqrt(2)v+1)

So the same integral can now be expressed using this identity as

2v2 dv / ((v2-sqrt(2)v+1)(v2+sqrt(2)v+1))

And now, breaking it down into partial fractions, the integral becomes,

(sqrt(2)/2) v dv / (v2-sqrt(2)v+1) + (-sqrt(2)/2) v dv / (v2+sqrt(2)v+1)

Now this is starting to get tricky, so I'll break the first term into two integrals I1 and I2, and the second term into two more integrals I3 and I4.

sqrt(tan(x)) dx is I1 + I2 + I3 + I4, where

I1 = (sqrt(2)/4)(2v-sqrt(2)) dv / (v2-sqrt(2)v+1)
I2 = (1/2) dv / (v2-sqrt(2)v+1)
I3 = (-sqrt(2)/4)(2v+sqrt(2)) dv / (v2+sqrt(2)v+1)
I4 = (1/2) dv / (v2+sqrt(2)v+1)

Now I'll do each of I1, I2, I3, and I4 separately:

I1 = (sqrt(2)/4)(2v-sqrt(2)) dv / (v2-sqrt(2)v+1)
This integral is of the form du/u, which is ln|u|, so
I1 = (sqrt(2)/4) ln|v2-sqrt(2)v+1| + C1

I2 = (1/2) dv / (v2-sqrt(2)v+1)
This can be converted into the form a/((av+b)2+1), if we let a=sqrt(2) and b=-1
I2 = (sqrt(2)/2) sqrt(2)/(2v2-2sqrt(2)v+1+1)
I2 = (sqrt(2)/2) sqrt(2)/((sqrt(2)v-1)2+1)
I2 = (sqrt(2)/2) atan(sqrt(2)v-1) + C2

I3 = (-sqrt(2)/4)(2v+sqrt(2)) dv / (v2+sqrt(2)v+1)
Again, this is the the ln form, so
I3 = (-sqrt(2)/4) ln(v2+sqrt(2)v+1) + C3

I4 = (1/2) dv / (v2+sqrt(2)v+1)
Again, this can be converted to the atan form, so
I4 = (sqrt(2)/2) atan(sqrt(2)v+1) + C4

To summarize,

I1 = (sqrt(2)/4) ln(v2-sqrt(2)v+1) + C1
I2 = (sqrt(2)/2) atan(sqrt(2)v-1) + C2
I3 = (-sqrt(2)/4) ln(v2+sqrt(2)v+1) + C3
I4 = (sqrt(2)/2) atan(sqrt(2)v+1) + C4

The sum of which gives us the final answer,

sqrt(tan(x)) =
(sqrt(2)/4) ln(v2-sqrt(2)v+1) + (sqrt(2)/2) atan(sqrt(2)v-1) + (-sqrt(2)/4) ln(v2+sqrt(2)v+1) + (sqrt(2)/2) atan(sqrt(2)v+1) + C =

(sqrt(2)/4) ln(tan(x)-sqrt(2tan(x))+1) + (sqrt(2)/2) atan(sqrt(2tan(x))-1) +
(-sqrt(2)/4) ln(tan(x)+sqrt(2tan(x))+1) + (sqrt(2)/2) atan(sqrt(2tan(x))+1) + C

Related pages in this website

Table of Integrals

Integration by Parts

Partial Fractions

 

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