Question

integration help

Answer 1

e

Answer 2

Right.

Answer 3

if u need to derive then take
x= tan u
dx = sec^2u du and so on...

Answer 4

care to explain the steps?

Answer 5

\(x = \sqrt{13} \tan u \implies dx = \sqrt{13} \sec ^2 u du\)

\[ \int \dfrac{1}{\sqrt{x^2+13}}dx = \int \sec u~ du \]

Answer 6

http://math2.org/math/integrals/more/sec.htm

Answer 7

ok, I'm not sure how you get to sqt tan u

Answer 8

do you mean the first step ?

Answer 9

yes

Answer 10

well, first notice that you cannot break a radical

Answer 11

you need to do some substitution and simplify it :

sin/cos will not work as we dont have any formulas like :
1+sin^2 = xx
or

1+cos^2 = xx

Answer 12

however 1+tan^2 = sec^2
so tan is the thing to substitute

Answer 13

ok

Answer 14

since you dont have just \(\sqrt{x^2+1}\), you need to fix ur substitution accordingly

Answer 15

and you use the Pythagoras theorem in the triangle so get the hypotenuse = sqrt (x^2 + 13)

using \[\tan \theta = \frac{ x }{ a }\]
\[x=\tan \theta \]

Answer 16

yes

\[

\int \dfrac{1}{\sqrt{x^2+13}}dx = \int \dfrac{1}{\sqrt{\dfrac{1}{13}\left(\dfrac{x^2}{13}+1\right)}}dx



\]

Answer 17

from above it should be convincing that \(\large \dfrac{x}{\sqrt{13}} = \tan u \) simplifies the radical

Answer 18

sorry, can you show me where 1/13 and x^2/13 + 1 came from?

Answer 19

sorry i made a mistake one sec let me correct it

Answer 20

\[\int \dfrac{1}{\sqrt{x^2+13}}dx = \int \dfrac{1}{\sqrt{13\left(\dfrac{x^2}{13}+1\right)}}dx

\]

Answer 21

see if that looks okay

Answer 22

sorry can you show me where the 13 and (x^2/13 + 1) comes from?

Answer 23

\[

\int \dfrac{1}{\sqrt{x^2+13}}dx = \int \dfrac{1}{\sqrt{\left(x^2*\dfrac{13}{13}+13\right)}}dx = \int \dfrac{1}{\sqrt{13\left(\dfrac{x^2}{13}+1\right)}}dx

\]

Answer 24

sorry, I'm just so lost cause I'm not very good at differentiation. Where did the 13/13 come from?

Answer 25

I'm not good at integration as well*

Answer 26

its okay, it got nothing to do with integrals however

Answer 27

its okay, it got nothing to do with integrals however

Answer 28

few more detailed steps :

\[

\dfrac{1}{\sqrt{x^2+13}} = \dfrac{1}{\sqrt{\left(x^2*1+13\right)}} = \dfrac{1}{\sqrt{\left(x^2*\dfrac{13}{13}+13\right)}} = \dfrac{1}{\sqrt{13\left(\dfrac{x^2}{13}+1\right)}}

\]

Answer 29

ok, I get what the maths side of what you did but how come you multiplied by 1 and where did it come from? and why does it turn into 13/13, can't you keep it as 1?

Answer 30

you can keep it as 1 but u will be stuck dead in waters there itself - your goal is to substitute something so that it simplifies the radical by getting a perfect square inside

Answer 31

You can argue that its kind of guessing- and you will be right ! the guessing comes by practice so dont wry right away, some day everything will make sense :P

Answer 32

ok..haha

Answer 33

watch this debate on integrals Vs derivatives for a good laugh :)
https://www.youtube.com/watch?v=iNtMLGvzFHA

Answer 34

woah it's about an hour long, I don't have that kind of time right now haha