Use substitution x=2tan(y) to integrate (37dx)/(x^2*sqrt(x^2+4)) in y

Consider that

[tex]\sqrt{4+x^2}=\sqrt{4+4\tan^2(y)}=2\sqrt{1+\tan^2(y)}[/tex]

and then use the approiate Pythagorean identity.

 

namely [tex]1+\tan^2(y)=\sec^2(y)[/tex] so that

[tex]2\sqrt{1+\tan^2(y)}=2\sqrt{\sec^2(y)}=2\sec (y)[/tex]

 

Also, for [tex]x=2\tan (y)[/tex] we have [tex]dx=2\sec^2 (y)dy[/tex]

 

I don't see how you could have gotten that!
Your integral is
[tex]\int\frac{37dx}{x^2\sqrt{x^2+4}}[/tex]
Letting x= 2 tan y, gives,as you were told before, dx= 2 sec²y dy, and [itex]\sqrt{x^2+4}= \sqrt{4tan^2y+4}= 2sec y[/itex]. Of course x²= 4tan² y. I don't see how you could have gotten that "-1" in "sec²y- 1".

 

Do you mean you changed that x² inthe denominator into
4tan² y and then into 4(sec²y-1)?? Why?

As was point out before, if x= 2 tan y, then dx= 2 sec² y
and x²+ 4= 4tan²y+ 4= 4 sec²y so that [itex]\sqrt{x^2y+ 4}= 2 sec y[/itex]. Your integral becomes
[tex]37\int \frac{2 sec^2 y dy}{(4tan^2 y)(2 sec y)}[/tex]
which reduces to
[tex]\frac{37}{4}\int\frac{sec y dy}{tan^2 y}[/tex]
Because of the odd power of sec y, I would be inclined to convert to sine and cosine now.