integral 3dx/sqrt(1+9x^2) I just need help with part of the algebra. .

Question

integral 3dx/sqrt(1+9x^2)      I just need help with part of the algebra. .

anonymous · Answer

$$\frac{ 3\sec^2\theta }{ \sqrt(1+9x^2) }$$

anonymous · Answer

dx is = 1/3 sec^2theta

anonymous · Answer

how do they get to sec^theta/sectheta

anonymous · Answer

more specifically, where does the 1/3 go.. and how do they get the 9 out of the radical while leaving the 1 there?

anonymous · Answer

$$\int\limits \frac{ 3\sec^2\theta }{ 3\sqrt{1+9x^2} }d \theta$$

anonymous · Answer

do you just cancel the 9 as a 3 and leave the sqrt 1 because sqrt 1 is 1 ?

anonymous · Answer

i guess i am getting confused at factoring the 9 out of the square root... would you have to divide the sqrt 1 by sqrt 9 as well when you factor it?  usually these problems have like 9+9x^2 in the denominator, thus why i am confused.

hartnn · Answer

i think you are getting confused because you kept your function in both x and theta 
keep your function in only theta and dtheta

anonymous · Answer

sloppy typing.. they are not mixed on my paper.

anonymous · Answer

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