Value of $ \int\frac{\sec x}{\sqrt{3+\tan x}}dx$

Question

Finding value of $\displaystyle \int\frac{\sec x}{\sqrt{3+\tan x}}dx$

Try: Let $\displaystyle I = \int \frac{\sec x}{\sqrt{3+\tan x}}dx,$ Put $3+\tan x = t^2$

and $\sec^2 xdx =2tdt$

So $\displaystyle I = 2\int \frac{1}{\sqrt{t^4-6t^2+10}}dt$

could some help me how to solve it, Thanks

You properly made it. The problem is that now you face elliptic integrals. — Claude Leibovici, Sep 13 '18 at 10:08
@ Claude Leibovici How can we express in Elleptical form. pleae explain. Thanks — DXT, Sep 13 '18 at 14:16
Enjoy the result https://www.wolframalpha.com/input?i=integrate+1%2Fsqrt(t%5E4-6t%5E2%2B10) — Claude Leibovici, Sep 13 '18 at 14:43

score 1 · Answer 1 · answered Sep 13 '18 at 18:31

1

Hint:

$$I=\int\frac{1}{\cos x \sqrt {3+ tan x}} dx= 2\cos x\frac{1}{2 \cos^2x \sqrt {3+ tan x}} dx$$

$$\frac{1}{2 \cos^2 x \sqrt {3+ tan x}} dx= du$$

$$u=\sqrt {3+ tan x}$$

$$v=2\cos x$$ $$dv=-2\sin x$$

$$I= \cos x \sqrt {3+ tan x} + 2\int \sin x \sqrt {3+ tan x} dx$$

Now take $3+ tan x=t^2 $ and continue.

answered Sep 13 '18 at 18:31

sirous

I am afraid that you get another elliptic integral. – Claude Leibovici Sep 14 '18 at 02:15

1 Answers1