Teacher needs help with an integral involving trig identities that a student found online

Question

$$\int\frac{\sqrt{\sec^3(x)+\tan^3(x)+\tan(x)}}{1+\sec(x)-\tan(x)}\,dx\quad\text{for $0<x<\frac{\pi}{2}$}$$

I have spent several days trying and failing to solve this problem. I am a teacher student of mine found it online. They asked for help solving it and initially I thought that there was a sign error because if so, I could easily solve it by substitution. However, the student maintains that it can be solved. I have tried so many things, I even resorted to using an online integral calculator which said there was no antiderivative. I am just so drained and demoralized, any help with this is appreciated.

Answer 1

Let $$I = \int \frac{\sqrt{\sec^3x+\tan^3x+\tan x}}{1+\sec x-\tan x}dx$$

Write the $1$ in the denominator as $\sec^2x-\tan^2x$, and substitute $$t = \sec x+\tan x$$ Or, $$dx=\frac{2dt}{t^2+1}$$

and we can write $\sec x = \frac{t^2+1}{2t}$ and $\tan x=\frac{t^2-1}{2t}$.

Substituting these in the integral, the integral now transform to

$$I= \int{\frac{t\sqrt{t \left(\frac{1}{t^2}+\frac{t^2+1}{2t}\frac{t^2-1}{2t} \right)+\frac{t^2-1}{2t}}}{t+1}}\frac{2dt}{t^2+1}$$

After some simplification, it simplifies to $$I=\int \frac{\sqrt t}{t+1}dt$$ which can be solved by taking $u=\sqrt{t}$, and the integral evaluates to $$I=2(\sqrt t-\arctan(\sqrt t))$$ and as, $t=\sec x+\tan x$, $$\boxed{I=2(\sqrt{\sec x+\tan x}-\arctan{\sqrt{\sec x+ \tan x})}+C}$$

Answer 2

I presume when you refer to sign error, you are refering to portion of $\sec(x)-\tan(x)$ in the denominator. If that's the case then just replace $\sec(x)-\tan(x)$ by $\dfrac{1}{\sec(x)+\tan(x)}$ and I believe you can solve the rest.

Edit: Just thought of including the full solution. $$\int \dfrac{\sqrt{\sec^3x + \tan^3x +\tan x}}{1+\sec x - \tan x}dx$$

$$= \int \dfrac{\sqrt{\sec^3x + \tan^3x +\tan x}}{1+\dfrac{1}{\sec x + \tan x}}dx $$ $$= \int \dfrac{(\sec x+ \tan x)\sqrt{\sec^3x + (\tan^2x +1)\tan x}}{1+\sec x + \tan x}dx $$ $$= \int \dfrac{\sec x(\sec x+ \tan x)\sqrt{\sec x +\tan x}}{1+\sec x + \tan x}dx $$

Now let, $$u=\sec x + \tan x + 1$$ $$\implies du = \sec x(\sec x + \tan x)dx$$ Hence our integral transforms into, $$\int \dfrac{\sqrt{u-1}}{u}du$$ Now again, substitute $u-1=z^2$ $$\implies du = 2z dz$$ Finally we have, $$\int \dfrac{2z^2}{z^2+1}dz$$

Note: This step could have been easily achieved by taking $u=\sec x + \tan x$ at the very beginning, but I decided to break this step down for easier understanding.

I leave it from here as this integral is very elementary. At the end we need to re-substitute $z$ in terms of $x$ to have our final answer.

Answer 3

The use of the $\sec$ is a bit confusing. The integral can be done with the standard substitution $u = \tan \frac{x}{2} $, but the result is terrible in terms of roots containing $ 2 + \sqrt 7$ and some others. The expression is to big to handle.

The given answer is more compact, but its hard to prove

D[2 (Sqrt[Sec[x] + Tan[x]] - ArcTan[Sqrt[Sec[x] + Tan[x]]]), x] // FullSimplify
(Sec[x] (Sec[x]+Tan[x])^(3/2))/(1+Sec[x]+Tan[x])

Change of sign in denominator and power $\dfrac12$ to $\dfrac32$

    Assuming[0<x<Pi/2,FullSimplify[ Sqrt[Sec[x]^3 + Tan[x]^3 + Tan[x]]/(1+ Sec[x]-Tan[x]) - 
( Sec[x] (Sec[x] + Tan[x])^(3/2))/(1 + Sec[x] + Tan[x]), {x, 0, 7}])

No result, but Series or plot do

     Series[ Sqrt[Sec[x]^3 + Tan[x]^3 + Tan[x]]/
(1 + Sec[x] - Tan[x]) ==
 (  Sec[x] (Sec[x] + Tan[x])^(3/2))/
(1 + Sec[x] + Tan[x]), {x, 0, 32}]
True