Indefinite Integral

lampat

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Dec 14, 2011
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1+tanxcsc2x+secx  dx\displaystyle \displaystyle \int\sqrt {\frac{1+\tan x}{\csc^2 x+\sqrt {\sec x}}}\,\,dx
 
1+tanxcsc2x+secx  dx\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.
 
My Try::

=1+tanxcsc2x+secxdx\displaystyle \displaystyle = \int \sqrt{\frac{1+\tan x}{\csc^2 x+\sqrt{\sec x}}dx}

=1+sinxcosx1sin2x+1cosxdx\displaystyle \displaystyle = \int \sqrt{\frac{1+\frac{\sin x}{\cos x}}{\frac{1}{\sin ^2 x}+\frac{1}{\sqrt{\cos x}}}}dx

=sinx+cosxcosxcosx+sin2xsin2x.cosxdx\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

=(sinx+cosx).sin2x(sin2x+cosx).cosxdx\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

=(sinx+cosx).sin2x(sin2x.cosx+cosx)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 cosx=t2\displaystyle \displaystyle \cos x = t^2 and sinxdx=2tdtdx=2t1t2\displaystyle \sin x dx = -2tdt \Leftrightarrow dx = \displaystyle -\frac{2t}{\sqrt{1-t^2}}

=(1t4+t2).(1t4)(t.(1t4))+t2.2t1t2dt\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
 
My Try::

1+tanxcsc2x+secxdx\displaystyle \displaystyle \int \sqrt{\frac{1+\tan x}{\csc^2 x+\sqrt{\sec x}}dx}

=1+sinxcosx1sin2x+1cosxdx\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

=sinx+cosxcosxcosx+sin2xsin2x.cosxdx\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

=(sinx+cosx).sin2x(sin2x+cosx).cosxdx\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

=(sinx+cosx).sin2x(sin2x.cosx+cosx)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....about as far as you can get with trig (?)
=sinx+cosx(cosx+cosx/sin2x)dx\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 x\displaystyle x.
To take square root, must have cosx0\displaystyle \cos{x} \ge 0, so π/2xπ/2\displaystyle -\pi/2 \le x \le \pi/2 .
Numerator sinx+cosx0    π/4x3π/4\displaystyle \sin{x} + \cos{x} \ge 0 \implies -\pi/4 \le x \le 3\pi/4.
Finally, can't have cosx=0\displaystyle \cos{x}=0 in the denominator, so xπ/2\displaystyle x \ne \pi/2.
The complete domain is π/4x<π/2\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 x\displaystyle x...

Let cosx=t2\displaystyle \displaystyle \cos x = t^2 and sinxdx=2tdtdx=2t1t2\displaystyle \sin x dx = -2tdt \Leftrightarrow dx = \displaystyle -\frac{2t}{\sqrt{1-t^2}}

=(1t4+t2).(1t4)(t.(1t4))+t2.2t1t2dt\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..
 
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