Question

Find the following indefinite integral:

Answer 1

\[\int\limits_{}^{} \frac{ \tan \sqrt{x} }{ \sqrt{x} } dx\]

Answer 2

\[\int\limits\limits_{}^{} \frac{ \tan \sqrt{x} }{ \sqrt{x} } dx \qquad = \qquad \int\limits\limits_{}^{} \tan \sqrt{x}\left(\frac{1}{\sqrt{x}} dx\right)\]See how I just took the tangent off of the top of the fraction?
From here you can do a `U substitution`, let \(u=\sqrt x\)

Answer 3

i was letting u=tanrootx

Answer 4

yah that's no good! :) heh

Answer 5

haha ok

Answer 6

so my du--1/x^2?

Answer 7

\[du=\frac{1}{2\sqrt x}\]You might want to convert your square root to a rational expression if you don't have this derivative memorized.
What I mean is, change it to x^1/2 so it's easier to differentiate.

Answer 8

woops I didn't write the dx on the right* it should be there.

Answer 9

okay so thats my du.

Answer 10

\[\large u=x^{1/2} \qquad \rightarrow \qquad du=\frac{1}{2}x^{-1/2}dx \qquad \rightarrow \qquad du=\frac{1}{2\sqrt x}\]Yup, there is our du.
See how the right side ALMOST matches the parthenthesis'd part in our integral?? Is there a way we can make them match, so we can apply the substitution?

Answer 11

I dropped the dx again!!! ahhhh

Answer 12

can we take out 1/2?

Answer 13

\[\large du=\frac{1}{2}\color{orangered}{\frac{1}{\sqrt x}dx}\]And here is our integral again,\[\int\limits\limits\limits_{}^{} \tan \sqrt{x}\left(\color{orangered}{\frac{1}{\sqrt{x}} dx}\right)\]

Answer 14

Take out? like on a date??
jk, yah i think you have the right idea maybe :)
you want to multiply both sides by 2, to get rid of that 1/2 on the right.

Answer 15

Ahhh sorry I have to go to school :C
If you're still stuck on this, i'll be @sirm3d or someone can help you finish it!! c:

Answer 16

i meant taking out a 1/2 outside the inegral sign haha.

Answer 17

thanks for your help!

Answer 18

after \[u=\sqrt x\]
square both sides
\[u^2=x\\2udu=dx\\\int \frac{\tan \sqrt x}{\sqrt x}dx=\int \frac{\tan u}{u}(2u\mathrm du)\]