Question

Evaluate the integral

Answer 1

\[\int\limits \frac{ 10 }{ x^2\sqrt{x^2+4} }\]

Answer 2

So for this one i am using x=atan theta

Answer 3

then i have x^2 = 2tan^2theta ?

Answer 4

because my a is 4, and 2 squared is 4

Answer 5

Your a is 2, yes?
Because 2 squared is 4.

Answer 6

\[\large\rm x=a \tan \theta\qquad\to\qquad x^2=a^2\tan^2\theta\]

Answer 7

yes ok thank you. and my dx = sec^2theta is that correct so far

Answer 8

\[\int\limits \frac{ 10 }{ 2\tan^2\theta \sqrt{2\tan^2\theta+4} }\sec^2\theta\]

Answer 9

Close, you gotta carry around all the little bits and pieces or you're going to miss things.\[\large\rm x=2\tan \theta\]\[\large\rm \color{royalblue}{dx=2\sec^2\theta~d \theta}\]\[\large\rm \color{orangered}{x^2=4\tan^2\theta}\]Do you see the 2 you missed in the dx?

Answer 10

\[\int\limits \frac{ 10 }{\color{red}4\tan^2\theta \sqrt{\color{red}4\tan^2\theta+4} }\sec^2\theta\]

Answer 11

\[\int\limits\limits \frac{ 10 }{\color{red}4\tan^2\theta \sqrt{\color{red}4\tan^2\theta+4} }\color{red}{2}\sec^2\theta\]

Answer 12

thank you! and ok are we squaring the 2 because of the a^2 and the x^2 is what adds a 2 in sec^2theta which originally was sec theta correct?

Answer 13

right!

Answer 14

hmm ok i am re doing the problem

Answer 15

ok i see now! you are finding dx and x^2 from x=tan theta. thanks!!

Answer 16

i believe i can use the identity 1+tan^2theta = sec^2 theta

Answer 17

but not sure what to do with the 4 in front of the tan^2theta. do i just put it in front of sec^2 theta? so 2sec^2theta ?

Answer 18

i mean the 4

Answer 19

Factor, yes.\[\large\rm 4\tan^2\theta+4=4(\tan^2\theta+1)=4(\sec^2\theta)\]

Answer 20

it would be \[4\sec^2\theta\] ...

Answer 21

ok! thanks

Answer 22

so after some steps i am left with \[\frac{ 5 }{ 4 } \int\limits \frac{ \sec \theta }{ \tan^2 \theta }\]

Answer 23

the way i got to that was :


\[\int\limits \frac{10}{4\tan^2\theta 4\sec \theta}2\sec^2 \theta\]

i multiplied the 10 and 2 to put it all in the numerator .. not sure if thats good?
\[\int\limits \frac{20\sec^2 \theta}{4\tan^2 \theta 4 \sec \theta}\]

so then i cancel out one of the sec^2 theta on top with the bottom and i divided the 4 and 20

and thats how i got the previous answer

Answer 24

Im thinking that isn't correct because I can't find anywhere to go from here.

Answer 25

mistake with the constants

Answer 26

\[\int\limits \frac{10}{4\tan^2\theta 2\sec \theta}2\sec^2 \theta\]

Answer 27

ok. i shouldn't have combined 2 and 10

Answer 28

oh and I should have taken the square root of 4, making it 2

Answer 29

\[\frac{5}{2}\int \frac{\sec(x)}{\tan^2(x)}dx\]\[\frac{5}{2}\int \frac{\sin(x)}{\cos^2(x)}dx\] a u-sub gets it now

Answer 30

how did you change sec x to sin x?

Answer 31

@satellite73 it should be cos / sin^2