Indefinite Integral

[lampat]

\(\displaystyle \displaystyle \int\sqrt {\frac{1+\tan x}{\csc^2 x+\sqrt {\sec x}}}\,\,dx\)

 

[DrPhil]

\(\displaystyle \displaystyle \int\sqrt {\frac{1+\tan x}{\csc^2 x+\sqrt {\sec x}}}\,\,dx\)

What have you done? We need to see your approach.

 

[lampat]

My Try::

\(\displaystyle \displaystyle = \int \sqrt{\frac{1+\tan x}{\csc^2 x+\sqrt{\sec x}}dx}\)

\(\displaystyle \displaystyle = \int \sqrt{\frac{1+\frac{\sin x}{\cos x}}{\frac{1}{\sin ^2 x}+\frac{1}{\sqrt{\cos x}}}}dx\)

\(\displaystyle \displaystyle = \int \sqrt{\frac{\frac{\sin x+\cos x}{\cos x}}{\frac{\sqrt{\cos x}+\sin^2 x}{\sin ^2 x.\sqrt{\cos x}}}}dx\)

\(\displaystyle \displaystyle =\int \sqrt{\frac{\left(\sin x+\cos x \right).\sin^2 x}{\left(\sin^2 x+\sqrt{\cos x}\right).\sqrt{\cos x}}}dx\)

\(\displaystyle \displaystyle = \int\sqrt{\frac{\left(\sin x+\cos x \right).\sin^2 x}{\left(\sin^2 x .\sqrt{\cos x}+\cos x \right)}}dx\)

Let \(\displaystyle \displaystyle \cos x = t^2 \) and \(\displaystyle \sin x dx = -2tdt \Leftrightarrow dx = \displaystyle -\frac{2t}{\sqrt{1-t^2}}\)

\(\displaystyle \displaystyle = -\int \sqrt{\frac{\left(\sqrt{1-t^4}+t^2 \right).\left(1-t^4\right)}{\left(t.\left(1-t^4\right)\right)+t^2}}.\frac{2t}{\sqrt{1-t^2}}dt\)

Now I have found this Complex Type expression,

would you like to explain me after that step,

Thanks

 

[DrPhil]

My Try::

\(\displaystyle \displaystyle \int \sqrt{\frac{1+\tan x}{\csc^2 x+\sqrt{\sec x}}dx}\)

\(\displaystyle \displaystyle = \int \sqrt{\frac{1+\frac{\sin x}{\cos x}}{\frac{1}{\sin ^2 x}+\frac{1}{\sqrt{\cos x}}}}dx\)....express everything as sine and cosine

\(\displaystyle \displaystyle = \int \sqrt{\frac{\frac{\sin x+\cos x}{\cos x}}{\frac{\sqrt{\cos x}+\sin^2 x}{\sin ^2 x.\sqrt{\cos x}}}}dx\)....numerator& denominator over LCDs

\(\displaystyle \displaystyle =\int \sqrt{\frac{\left(\sin x+\cos x \right).\sin^2 x}{\left(\sin^2 x+\sqrt{\cos x}\right).\sqrt{\cos x}}}dx\)....invert denominators

\(\displaystyle \displaystyle = \int\sqrt{\frac{\left(\sin x+\cos x \right).\sin^2 x}{\left(\sin^2 x .\sqrt{\cos x}+\cos x \right)}}dx\)....about as far as you can get with trig (?)

\(\displaystyle \displaystyle = \int\sqrt{\frac{\sin x+\cos x }{\left( \sqrt{\cos x}+\cos x/ \sin^2x \right)}}dx\)....is that any help?

I would like to pause here and consider the domain of \(\displaystyle x\).
To take square root, must have \(\displaystyle \cos{x} \ge 0\), so \(\displaystyle -\pi/2 \le x \le \pi/2 \).
Numerator \(\displaystyle \sin{x} + \cos{x} \ge 0 \implies -\pi/4 \le x \le 3\pi/4\).
Finally, can't have \(\displaystyle \cos{x}=0\) in the denominator, so \(\displaystyle x \ne \pi/2\).
The complete domain is \(\displaystyle -\pi/4 \le x < \pi/2\).

Since the domain includes points in both 4th and 1st quadrants, the cosine may not the best choice for substitution because it is not 1:1 with \(\displaystyle x\)...

Let \(\displaystyle \displaystyle \cos x = t^2 \) and \(\displaystyle \sin x dx = -2tdt \Leftrightarrow dx = \displaystyle -\frac{2t}{\sqrt{1-t^2}}\)

\(\displaystyle \displaystyle = -\int \sqrt{\frac{\left(\sqrt{1-t^4}+t^2 \right).\left(1-t^4\right)}{\left(t.\left(1-t^4\right)\right)+t^2}}.\frac{2t}{\sqrt{1-t^2}}dt\)

Now I have found this Complex Type expression,

would you like to explain me after that step,

Thanks

I was very impressed by the algebraic tricks you used on the max(x^4y + xy^4) question, using the constraint (x+y)=1 to simplify the functional form. Can you do something like that here?? I will keep looking..