Question

Can someone help me with this ingetral. I will write it out.

Answer 1

\[\int\limits_{}^{} (\sec^2\sqrt{x})/\sqrt{x}\]

Answer 2

I get as far as letting u= \[\sqrt{x}\]

Answer 3

thats right. let:
\[u = \sqrt{x} \iff du = \frac{1}{2\sqrt{x}}dx\]

When you plug that back into the integral you should get:
\[2\int\limits \sec^2(u)du\]

Answer 4

where does that 2 outside of the integral come from?

Answer 5

or rather what does, dx=

Answer 6

it comes from:
\[ \int\limits \frac{\sec^2(\sqrt{x})}{\sqrt{x}}dx = \int\limits \sec^2(\sqrt{x})*\left(\frac{1}{\sqrt{x}}dx\right)\]

from my last post:
\[u = \sqrt{x} \iff du = \frac{1}{2\sqrt{x}}dx \iff 2du = \frac{1}{\sqrt{x}}dx\]

Answer 7

sp the u substitution gives:
\[\int\limits \sec^2(u) (2du) = 2\int\limits \sec^2(u)du\]

Answer 8

so*

Answer 9

but how is it that we get he 2 to be in the numerator

Answer 10

oh we multiplied both sides by 2

Answer 11

right.

Answer 12

but shouldnt the bottom part of the intetgral also get subbed with a U since it is sqrt x

Answer 13

oh i see

Answer 14

when i solve the derivaive of U for dex i get 2(sqrt x) du

Answer 15

which means that the bottom sqrt x gets divided out

Answer 16

right, and if you plug that in stuff cancels and you end up with the clean integral with u's, and a 2 on the outside.

Answer 17

you da best, man

Answer 18

thank you :)

Answer 19

i started working on this problem last night, and i couldnt sleep becuase i couldnt figure it. Thats why i am up at 3 am in the morining. But now i see i wasnt actually trying hard enough

Answer 20

it just take practice.

practice practice practice.