Integration of squareroot tanx dx?
Solution:
𝑊𝑒 𝑛𝑒𝑒𝑑 𝑡𝑜 𝑓𝑖𝑛𝑑 ∫ √tan x dx
First, we can make a u − substitution. Let u = √tan x. Then du =
sec2 xdx
2√tan x
dx
= 2 cos2 x
√tan x
.
Recall , cos2 x + sin2 x = 1. If we factor out a cos 2 x on the left side, we get (cos2 x)(1 +
tan2 x) = 1. Now we can rewrite the integral:
√tan x
∫ √tan x dx = ∫
dx
(cos 2 x)(1 + tan2 x)
Now, in integrand we multiply the nominator and the denominator by 2√tan x. After doing this,
we get:
2 tan x
∫ √tan x dx = ∫
dx
(2cos2 x)√tan x (1 + tan2 x)
2 tan x = 2u2 and 1 + tan2 x = 1 + u4 . If we rewrite the integrand in terms of variable u, we
get:
2u2
∫ √tan x dx = ∫
du
1 + u4
Then, we can factor 1 + u4 as:
1 + u4 = (u2 + u√2 + 1)(u2 − u√2 + 1);
Now we can rewrite the integral:
2u2
∫ √tan x dx = ∫
du
(u2 + u√2 + 1)(u2 − u√2 + 1)
Now, we have to use partial fractions:
2u2
Au + B
Cu + D
=
+
(u2 + u√2 + 1)(u2 − u√2 + 1) u2 + u√2 + 1 u2 − u√2 + 1
In the right side we reduce terms to a common denominator. Then we go through and equate
the coefficients of each power of u in the left side and in the right side so that we can solve
for A, B, C, and D.
2u2
(u2 + u√2 + 1)(u2 − u√2 + 1)
⟹
2u2
(u2 + u√2 + 1)(u2 − u√2 + 1)
=
=
Au + B
u2 + u√2 + 1
+
Cu + D
u2 − u√2 + 1
(Au + B)(u2 − u√2 + 1) + (Cu + D)(u2 + u√2 + 1)
(u2 + u√2 + 1)(u2 − u√2 + 1)
⟹ 2u2 = (Au + B)(u2 − u√2 + 1) + (Cu + D)(u2 + u√2 + 1)
⟹ 2u2 = Au3 − Au2 √2 + Au + Bu2 − Bu√2 + B + Cu3 + Cu2 √2 + Cu + Du2 + Du√2 + D
⟹ 2u2 = (A + C)u3 + (−A√2 + B + C√2 + D)u2 + (A − B√2 + C + D√2)u + (B + D)
Equations derived from coefficients near similar terms:
A + C = 0(I)
−A√2 + B + C√2 + D = 2 (II)
A − B√2 + C + D√2 = 0 (III)
B+D=0
(IV)
{
Equations (I) and (IV) tell us that C = −A and D = −B. Using that information, we replace the
expressions in equations (II) and (III). This gives us the following:
−A√2 + B + (−A)√2 + (−B) = 2 ⟹ A = −
√2
2
A − B√2 + (−A) + (−B)√2 = 0 ⟹ B = 0
So, we see that C =
√2
and
2
D = 0. Now, we go back to our partial fractions and fill in A, B, C,
and D. Doing this, we have the following:
2
2u
(u2 + u√2 + 1)(u2 − u√2 + 1)
=
√2
−( 2 )u + 0
u2 + u√2 + 1
+
√2
( 2 )u + 0
u2 − u√2 + 1
=
√2
√2
− 2 u
2 u
=
+
u2 + u√2 + 1 u2 − u√2 + 1
Finally, returning to the original problem, we can rewrite the integral as follows:
√2
√2
u
u
2
2
∫ √tan x dx = ∫(
+
)du
u2 + u√2 + 1 u2 − u√2 + 1
Now, we can break ∫ √tan x dx into two different integrals. Also, we can rewrite the numerator
of both integrals. Doing so will allow us to continue our integrations.
−
Notice that
√2
u
2
=
√2
(2u
4
1
− √2) + 2 and likewise that −
√2
u
2
=−
√2
(2u
4
1
+ √2) + 2 . The
reason for this will become evident shortly. Here is the first part of the integration, broken into
two separate pieces.
−
√2
u
2
∫ u2 +u√2+1 du = ∫
√2
(2u+√2)
4
u2 +u√2+1
−
1
2
du + ∫ u2 +u√2+1 du
Now, let us consider the first of these two integrals: ∫
√2
(2u+√2)
4
u2 +u√2+1
−
du
Here, the substitution is actually straightforward because of the above work. Let ν = u2 +
+u√2 + 1. Then that means that dν = (2u + √2)du, which is precisely what we have above.
(The constant can be factored out.) That means that we have:
∫
√2
(2u+√2)
4
u2 +u√2+1
−
du = −
√2 dν
∫ν
4
=−
√2
ln|ν|
4
+ C1 = −
√2
ln|u2
4
+ u√2 + 1| + C1
1
2
Second integral: ∫ u2 +u√2+1
du
1
1
1
1
2
=
=
=
=
u2 + u√2 + 1 2(u2 + u√2 + 1) 2u2 + 2u√2 + 2 2u2 + 2u√2 + 1 + 1
=
1
2
(u√2 + 1) + 1
=
1
√2
∙
2
√2 (u√2 + 1) + 1
=
√2
√2
∙
2 (u√2 + 1)2 + 1
That means that we have the following:
1
√2
√2
2
∫
𝑑𝑢 =
∫
𝑑𝑢
2 (𝑢√2 + 1)2 + 1
𝑢2 + 𝑢√2 + 1
We can finally integrate further, if we use another substitution. Let 𝜈 = 𝑢√2 + 1, then
𝑑𝜈 = √2𝑑𝑢 . Using that substitution, we have:
√2
√2
𝑑𝑢
∫
2
2
(𝑢√2+1) +1
=
𝑑𝜈 √2
√2
∫ 𝜈2 +1= 2 𝑡𝑎𝑛−1 𝑣
2
+ 𝐶2 .
Evaluating this, we have
1
√2
2
∫
𝑑𝑢 =
𝑡𝑎𝑛−1 (𝑢√2 + 1) + 𝐶2
2
2
𝑢 + 𝑢√2 + 1
Putting this all together, we have done half of the integration. So far, we have shown:
√2
− 2 𝑢
√2
√2
∫
= − (−
𝑙𝑛|𝑢2 + 𝑢√2 + 1| + 𝐶1 ) + ( 𝑡𝑎𝑛−1(𝑢√2 + 1) + 𝐶2 )
4
2
𝑢2 + 𝑢√2 + 1
√2
𝑢
2
√2
𝑢
2
Now, thankfully, something similar happens for ∫ 𝑢2 −𝑢√2+1 𝑑𝑢.We rewrite ∫ 𝑢2 −𝑢√2+1 𝑑𝑢
√2
(2𝑢−√2)
1
2
as ∫ 𝑢42 −𝑢√2+1 𝑑𝑢 + ∫ 𝑢2 −𝑢√2+1
𝑑𝑢.
For the first integral, we again use a substitution. Let 𝜈 = 𝑢2 − 𝑢√2 + 1. Then that means that
𝑑𝜈 = (2𝑢 − √2)𝑑𝑢 , which is precisely what we have earlier. (The constant can be factored
out.) That means that we have:
√2
(2𝑢 − √2)
√2 𝑑𝜈 √2
√2
∫ 4
𝑑𝑢 =
∫
=
𝑙𝑛|𝜈| + 𝐶3 =
𝑙𝑛|𝑢2 − 𝑢√2 + 1| + 𝐶3
2
4
𝜈
4
4
𝑢 − 𝑢√2 + 1
1
2
And for the second integral, we again use a trick to change ∫ 𝑢2 −𝑢√2+1
𝑑𝑢 into something
more manageable. Following the steps from above, we see that:
1
√2
√2
2
∫
𝑑𝑢 =
∫
𝑑𝑢
2 (𝑢√2 − 1)2 + 1
𝑢2 − 𝑢√2 + 1
Again, we use a substitution. Let 𝜈 = 𝑢√2 − 1. Then that means that 𝑑𝜈 = √2𝑑𝑢. That means
that we have:
1
√2
2
∫
𝑑𝑢 =
𝑡𝑎𝑛−1 (𝑢√2 − 1) + 𝐶4
2
2
𝑢 − 𝑢√2 + 1
That means we have the following:
√2
√2
√2
2 𝑢
∫
𝑑𝑢 = ( 𝑙𝑛|𝑢2 − 𝑢√2 + 1| + 𝐶3 ) + ( 𝑡𝑎𝑛−1(𝑢√2 − 1) + 𝐶4 )
4
2
𝑢2 − 𝑢√2 + 1
And so, putting together everything from above, we see that:
∫ √𝑡𝑎𝑛 𝑥 𝑑𝑥 = − (−
√2
√2
𝑙𝑛|𝑢2 + 𝑢√2 + 1| + 𝐶1 ) + ( 𝑡𝑎𝑛−1 (𝑢√2 + 1) + 𝐶2 ) +
4
2
√2
√2
+ ( 𝑙𝑛|𝑢2 − 𝑢√2 + 1| + 𝐶3 ) + ( 𝑡𝑎𝑛−1 (𝑢√2 − 1) + 𝐶4 )
4
2
Now, we can simplify the above expression just a bit. We can factor
√2
2
out of all terms. Also, we
can use a property of logs to simplify the two natural log expressions. Doing so, we get the
following expression:
1
𝑢2 − 𝑢√2 + 1
√2
−1
−1
∫ √𝑡𝑎𝑛 𝑥 𝑑𝑥 =
(𝑡𝑎𝑛 (𝑢√2 + 1) + 𝑡𝑎𝑛 (𝑢√2 − 1) + 𝑙𝑛 |
|) + 𝐶
2
2
𝑢2 + 𝑢√2 + 1
Lastly, we replace u with its expression from the first page. Doing this, we get our final answer:
∫ √𝑡𝑎𝑛 𝑥 𝑑𝑥 =
√2
(𝑡𝑎𝑛−1(√2 𝑡𝑎𝑛 𝑥
2
1
𝑡𝑎𝑛 𝑥−√2 𝑡𝑎𝑛 𝑥+1
1
𝑡𝑎𝑛 x−√2 tan x+1
+ 1) + 𝑡𝑎𝑛−1 (√2 𝑡𝑎𝑛 𝑥 − 1) + 2 𝑙𝑛 |𝑡𝑎𝑛 𝑥+√2 𝑡𝑎𝑛 𝑥+1|) + 𝐶,
where C is an arbitrary real constant.
Answer:
∫ √𝑡𝑎𝑛 𝑥 𝑑𝑥 =
√2
(𝑡𝑎𝑛−1(√2 𝑡𝑎𝑛 𝑥
2
+ 1) + 𝑡𝑎𝑛−1 (√2 𝑡𝑎𝑛 𝑥 − 1) + 2 𝑙𝑛 | tan x+√2 tan x+1 |) + C.