David Skrovanek
University of Pittsburgh
Evaluating ∫ √𝑡𝑎𝑛𝑥 𝑑𝑥
This intimidating integral can be solved by three distinct techniques. The first method I will describe is the most
elegant, but requires more “trickery”. The second method is more straightforward, but again demands some clever
manipulation. The third method is characterized by pure brute force.
Method 1:
Let T= ∫ √𝑡𝑎𝑛𝑥 𝑑𝑥 and
C=∫ √𝑐𝑜𝑡𝑥 𝑑𝑥
𝑠𝑖𝑛𝑥
𝑐𝑜𝑠𝑥
T + C = ∫ √𝑡𝑎𝑛𝑥 𝑑𝑥 + √𝑐𝑜𝑡𝑥 𝑑𝑥 = ∫ √𝑐𝑜𝑠𝑥 + √ 𝑠𝑖𝑛𝑥 𝑑𝑥
Now obtain a common denominator, and use sin 2𝑥 = 2𝑐𝑜𝑠𝑥𝑠𝑖𝑛𝑥:
𝑠𝑖𝑛𝑥
𝑠𝑖𝑛𝑥
𝑐𝑜𝑠𝑥
𝑐𝑜𝑠𝑥
𝑠𝑖𝑛𝑥 + 𝑐𝑜𝑠𝑥
𝑠𝑖𝑛𝑥 + 𝑐𝑜𝑠𝑥
𝑠𝑖𝑛𝑥 + 𝑐𝑜𝑠𝑥
√
∗√
+ √
∗√
=
=
= √2 (
)
𝑐𝑜𝑠𝑥
𝑠𝑖𝑛𝑥
𝑠𝑖𝑛𝑥
𝑐𝑜𝑠𝑥 √𝑐𝑜𝑠𝑥 ∗ 𝑠𝑖𝑛𝑥
√𝑠𝑖𝑛2𝑥
√𝑠𝑖𝑛2𝑥
2
T + C = √2 ∫
𝑠𝑖𝑛𝑥+𝑐𝑜𝑠𝑥
√𝑠𝑖𝑛2𝑥
𝑑𝑥
Recognize that (𝑠𝑖𝑛𝑥 − 𝑐𝑜𝑠𝑥 )2 = 1 − 𝑠𝑖𝑛2𝑥
Let 𝑢 = 𝑠𝑖𝑛𝑥 − 𝑐𝑜𝑠𝑥
𝑑𝑢 = (𝑠𝑖𝑛𝑥 + 𝑐𝑜𝑠𝑥 )𝑑𝑥
T + C =√2 ∫
𝑠𝑖𝑛𝑥+𝑐𝑜𝑠𝑥
√1−(𝑠𝑖𝑛𝑥−𝑐𝑜𝑠𝑥)
𝑑𝑥 = √2 ∫
2
𝑑𝑢
√1−𝑢2
= √2 sin−1 𝑢 = √2 sin−1 (𝑠𝑖𝑛𝑥 − 𝑐𝑜𝑠𝑥 )
Now we evaluate T - C= ∫ √𝑡𝑎𝑛𝑥 − √𝑐𝑜𝑡𝑥 𝑑𝑥
Using similar techniques as done for T + C, we obtain
√2 ∫
𝑠𝑖𝑛𝑥 − 𝑐𝑜𝑠𝑥
√𝑠𝑖𝑛2𝑥
𝑑𝑥 = √2 ∫
𝑠𝑖𝑛𝑥 − 𝑐𝑜𝑠𝑥
√(𝑠𝑖𝑛𝑥 + 𝑐𝑜𝑠𝑥)2 − 1
𝑑𝑥
Let 𝑢 = 𝑠𝑖𝑛𝑥 + 𝑐𝑜𝑠𝑥
𝑑𝑢 = (𝑐𝑜𝑠𝑥 − 𝑠𝑖𝑛𝑥 )𝑑𝑥 = −(𝑠𝑖𝑛𝑥 − 𝑐𝑜𝑠𝑥 )𝑑𝑥
T + C = −√2 ∫
𝑑𝑢
√𝑢2 −1
= −√2 cosh−1 𝑢 = −√2 cosh−1 (𝑠𝑖𝑛𝑥 + 𝑐𝑜𝑠𝑥)
Now to isolate T with a little algebra,
David Skrovanek
University of Pittsburgh
𝑇=
(𝑇−𝐶 )+(𝑇+𝐶)
2
=
√2
[sin−1(𝑠𝑖𝑛𝑥
2
− 𝑐𝑜𝑠𝑥) − cosh−1(𝑠𝑖𝑛𝑥 + 𝑐𝑜𝑠𝑥)]
Method 2:
∫ √𝑡𝑎𝑛𝑥 𝑑𝑥
𝑢2 = 𝑡𝑎𝑛𝑥
2𝑢 𝑑𝑢 = 𝑠𝑒𝑐2𝑥 𝑑𝑥
2𝑢 𝑑𝑢 = 1 + 𝑡𝑎𝑛2𝑥 𝑑𝑥
2𝑢 𝑑𝑢 = 1 + (𝑢2 )2 𝑑𝑥
2𝑢 𝑑𝑢 = 1 + 𝑢4 𝑑𝑥
∫
2𝑢2
𝑑𝑢
1 + 𝑢4
You can either use partial fractions (method 3), or be a little sneaky.
Here is the sneaky way:
∫
(𝑢2 + 1) + (𝑢2 − 1)
𝑑𝑢
1 + 𝑢4
Now we can break this up into two separate integrals. I’ll call them I1 and I2.
I1:
𝑢2 + 1
∫ 4
𝑑𝑢
𝑢 +1
Now let’s divide everything by u2.
1
1+ 2
𝑢
𝐼1 = ∫
𝑑𝑢
1
𝑢2 + 2
𝑢
Unlike normal u-substitution where one declares u, I am doing to define the differential dz first and then make my
z.
1
du
𝑢2
1
−
𝑢
dz=1 +
z=𝑢
If we square z we get something very nice…
1
−2
𝑢2
1
𝑧 2 + 2 = 𝑢2 + 2
𝑢
𝑧 2 = 𝑢2 +
Now we can integrate this.
David Skrovanek
University of Pittsburgh
𝐼1 = ∫
1
𝑑𝑧
𝑧2 + 2
This is an easy trig-sub problem (use tangent substitution).
1
𝐼1 =
tan−1
√2
𝑧
√2
Plugging in z and then u:
1
𝐼1 =
tan−1
𝑡𝑎𝑛𝑥 − 1
√2
√2𝑡𝑎𝑛𝑥
Now let’s integrate I2.
We will do the same steps we did for I1, by dividing by u2 and then defining dz. Here’s what it looks like in terms
of z.
1
𝐼2 = ∫ 2
𝑑𝑧
𝑧 −2
You can either factor this into (𝑧 + √2)(𝑧 − √2) and use partial fractions, or you can use trig-substitution (using
secant substitution). Both are tedious, but I would personally use partial fractions, as it was easy to factor.
𝐼2 =
1
2√2
ln
𝑧 − √2
𝑧 + √2
Plug in z and u to get I2.
𝐼2 =
1
2√2
ln
tan 𝑥 − √2 tan 𝑥 + 1
tan 𝑥 + √2 tan 𝑥 + 1
Finally, I1+I2 gives the answer.
1
√2
tan−1
𝑡𝑎𝑛𝑥 − 1
√2𝑡𝑎𝑛𝑥
+
1
2√2
ln
tan 𝑥 − √2 tan 𝑥 + 1
tan 𝑥 + √2 tan 𝑥 + 1
Method 3:
In the same manner as method 2, obtain the following:
∫
2𝑢2
𝑢2
𝑑𝑢
𝑜𝑟
2
∫
𝑑𝑢
1 + 𝑢4
1 + 𝑢4
First factor the u4+1 into (𝑢2 + √2𝑢 + 1)(𝑢2 − √2𝑢 + 1). This becomes a tedious partial fractions problem.
Most people would then use classical methods to decompose the fraction in the following manner:
𝐴𝑢 + 𝐵
𝑢2 + √2𝑢 + 1
+
𝐶𝑢 + 𝐷
𝑢2 − √2𝑢 + 1
=
2𝑢2
1 + 𝑢4
Doing it this way leads to a nasty system of equations that perhaps cannot be solved without the aid of linear
algebra. This yields the following system of equations:
David Skrovanek
University of Pittsburgh
1
0
−√2 1
[
1 −√2
0
1
1
√2
1
0
0
𝐴
0
1
𝐵
] ∗ [ ] = (2)
𝐶
0
√2
0
𝐷
1
Using [𝑥] = 𝐴−1 𝐵, (I will leave finding the inverse matrix to the reader)
𝐴=
−√2
√2
, 𝐵 = 0, 𝐶 =
,𝐷 = 0
2
2
However, this can be solved more elegantly with the aid of complex analysis. We will decompose the fraction in
the same manner one would compute residues.
First, we must further factor the denominator.
𝑢2 + √2𝑢 + 1 = (𝑢 −
𝑖√2 √2
𝑖√2 √2
+ )(𝑢 +
+ )
2
2
2
2
𝑢2 − √2𝑢 + 1 = (𝑢 −
𝑖√2 √2
𝑖√2 √2
− )(𝑢 +
− )
2
2
2
2
The expression can now be written as
(1)
(2)
(3)
(4)
𝑢2
𝐴0
𝐴0
𝐴0
𝐴0
=
+
+
+
1 + 𝑢4
𝑖√2 √2
𝑖√2 √2
𝑖√2 √2
𝑖√2 √2
𝑢−
+
𝑢+
+
𝑢−
−
𝑢+
−
2
2
2
2
2
2
2
2
To compute each 𝐴𝑛0 , simply employ the technique to compute residues of simple poles. One can either use the
𝑃(𝑥)
shortcut 𝑅𝑒𝑠(𝐴𝑛0 ) =
or simply take the limit as x approaches the respective pole. To illustrate:
𝑄′(𝑥)
(1)
𝐴0 =
𝑢2
lim
𝑖√2 √2
𝑢→ 2 − 2
(𝑢2 − √2𝑢 + 1) (𝑢 +
(2)
𝐴0 =
−√2 𝑖√2
−
8
8
−√2 𝑖√2
+
8
8
(3)
𝐴0 =
(4)
𝑖√2 √2
+ )
2
2
=
𝐴0 =
√2 𝑖√2
−
8
8
√2 𝑖√2
+
8
8
(𝑛)
Now comes the most tedious part—adding together each 𝐴0 .
√2
√2
√2
√2
(−1 − 𝑖)
(−1 + 𝑖)
(1 − 𝑖)
(1 + 𝑖)
𝑢2
8
8
8
8
=
+
+
+
1 + 𝑢4
𝑖√2 √2
𝑖√2 √2
𝑖√2 √2
𝑖√2 √2
𝑢−
+
𝑢+
+
𝑢−
−
𝑢+
−
2
2
2
2
2
2
2
2
David Skrovanek
University of Pittsburgh
𝑢2
𝑢
𝑢
√2
√2
=
(
)−
(
)
4
1+𝑢
4 𝑢2 − √2𝑢 + 1
4 𝑢2 + √2𝑢 + 1
Doing the correct algebra and multiplying by two yields:
√2𝑢
√2𝑢
∫
𝑑𝑢 − ∫
𝑑𝑢
2(𝑢2 − √2𝑢 + 1)
2(𝑢2 + √2𝑢 + 1)
After getting this, you have to use trig substitution on both integrals, but this becomes tricky as you must first
complete the square on both. For the first integral, the denominator becomes (𝑥 −
integral has (𝑥 +
√2 2
)
2
√2
𝑡𝑎𝑛𝜃 = 𝑢 −
+
1
2
and the second
1
2
+ .
∫
1
√2 2
)
2
1
√2
,
2
√2
√2𝑢
2(𝑢2 − √2𝑢 + 1)
𝑠𝑒𝑐𝜃 = √𝑢2 − √2𝑢 + 1,
𝑑𝑢
𝑑𝑢 =
1
√2
𝑠𝑒𝑐𝜃 2 𝑑𝜃,
𝑡𝑎𝑛𝜃 = √2𝑢 − 1
1
1
√2
𝑠𝑒𝑐𝜃 2 ∗ ( 𝑡𝑎𝑛𝜃 + ) 𝑑𝜃
2
√2 √2
√2
√2
√2
√2
√2
∫
=∫
𝑡𝑎𝑛𝜃 𝑑𝜃 + ∫
𝑑𝜃 =
ln 𝑠𝑒𝑐𝜃 +
𝜃
1
2
2
2
2
2
𝑠𝑒𝑐𝜃 2
2
=
√𝑢2 − √2𝑢 + 1
√2
√2
ln |
|+
tan−1(√2𝑢 − 1)
2
2
√2
𝑥
ln(2)
√2
2
Use the fact that 𝑙𝑛 ( ) = ln(𝑥) −
=
and the power identity of logarithms
√2
√2
( ln|𝑢2 − √2𝑢 + 1| − 𝑙𝑛2) +
tan−1 (√2𝑢 − 1)
4
2
Doing similar techniques for the other integral yields:
−∫
√2𝑢
2(𝑢2 + √2𝑢 + 1)
𝑑𝑢 = −
√2
√2
( ln|𝑢2 + √2𝑢 + 1| − 𝑙𝑛2) +
tan−1 (√2𝑢 + 1)
4
2
Combining, using the subtractive property of logarithms, we obtain
𝑢2 − √2𝑢 + 1
√2
√2
ln |
|+
[tan−1(√2𝑢 − 1) + tan−1 (√2𝑢 + 1)]
2
4
2
𝑢 + √2𝑢 + 1
In order to make this in the form of method 2’s answer, we must employ some trig identities for arctan.
tan−1 (−𝑥) = − tan−1 (𝑥)
David Skrovanek
University of Pittsburgh
1
𝜋
tan−1 ( ) = − tan−1 (𝑥) ±
𝑥
2
𝑢
+
𝑣
tan−1 𝑢 + tan−1 𝑣 = tan−1 (
)
1 − 𝑢𝑣
√2
[tan−1(√2𝑢 − 1) + tan−1(√2𝑢 + 1)]
2
2√2𝑢
𝑢2 − 1
√2
√2
−1 ( √2𝑢 ) = √2 tan−1 (
=
tan−1 (
)
=
−
tan
)
2
2 − 2𝑢2
2
𝑢2 − 1
2
√2𝑢
Finally plugging in our substitution, 𝑢 = √𝑡𝑎𝑛𝑥 and using the fact that
1
√2
tan−1
𝑡𝑎𝑛𝑥 − 1
√2𝑡𝑎𝑛𝑥
+
1
2√2
ln
√2
4
tan 𝑥 − √2 tan 𝑥 + 1
tan 𝑥 + √2 tan 𝑥 + 1
just as method 2 proved. Phew!
=
1
2√2
𝑎𝑛𝑑
√2
2
=
1
√2
,