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

[beanryu]

I feel so doing this problem.
PLEASE HELP! AND TEACH ME How TO DO IT.
My question is this.

use the substitution x=2tan(y)

(37dx)/(x^2*sqrt(x^2+4))

give the answer in terms of y.

I did the substition, but it looked more complicated.
It doesn't look like a u*du thing, but it does look like some arcsin thing... i couldn't figure it out. The x^2 is bugging me.
PLease help!

 

[benorin]

Consider that

$$\sqrt{4+x^2}=\sqrt{4+4\tan^2(y)}=2\sqrt{1+\tan^2(y)}$$

and then use the approiate Pythagorean identity.

 

[benorin]

namely $$1+\tan^2(y)=\sec^2(y)$$ so that

$$2\sqrt{1+\tan^2(y)}=2\sqrt{\sec^2(y)}=2\sec (y)$$

 

[benorin]

Also, for $$x=2\tan (y)$$ we have $$dx=2\sec^2 (y)dy$$

 

[beanryu]

i got

original expression= (37/4)*((sec(y))/((sec(y))^2-1)*dy

but I am stuck.

I tried to do it so that its integral is ln(u)*du, but then sec's derivative would come up...

 

[HallsofIvy]

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

 

[beanryu]

1+(tany)^2=(secy)^2
(tany)^2=(secy)^2-1

4(tany)^2=4((secy)^2-1)

this is how i got it... what's wrong?

 

[HallsofIvy]

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 $\sqrt{x^2y+ 4}= 2 sec y$. Your integral becomes
$$37\int \frac{2 sec^2 y dy}{(4tan^2 y)(2 sec y)}$$
which reduces to
$$\frac{37}{4}\int\frac{sec y dy}{tan^2 y}$$
Because of the odd power of sec y, I would be inclined to convert to sine and cosine now.

 

[beanryu]

thanx for the guide! I did it.