Question

How do I integrate this function using an inverse trig function?

Answer 1

\[f(x)=\frac{ -30 }{ \sqrt{1-25x^2} }\]

Answer 2

that is the function to integrate

Answer 3

Let's recall the arcsine integral:\[\Large \int\limits\limits \frac{1}{\sqrt{1-x^2}}dx \quad=\quad \arcsin x\]

Answer 4

So let's try to make our function match that form.

Answer 5

\[\Large \int\limits \frac{-30}{\sqrt{1-(5x)^2}}dx\]

Answer 6

Do you understand what I did on that step? I brought the 25 into the square.

Answer 7

I guess we can also bring the -30 out of the integral.\[\Large -30\int\limits\limits \frac{1}{\sqrt{1-(5x)^2}}dx\]

Answer 8

Make the substitution \(\Large u=5x\)

Answer 9

ok that makes sense

Answer 10

If we do that, what do we get in place of our dx?

Answer 11

5?

Answer 12

Well yes, we get 5 on that side of the equation,\[\Large du=5dx\]But we want to `replace` the dx in the problem, so we'll need to solve for dx here.
Looks like we'll divide both sides by 5, yes?\[\Large \frac{1}{5}du=dx\]

Answer 13

yes

Answer 14

Ok plugging in our substitution:\[\Large -30\int\limits\limits\limits \frac{1}{\sqrt{1-(5x)^2}}dx \qquad=\qquad -30\int\limits\frac{1}{1-u^2}\left(\frac{1}{5}du\right)\]

Answer 15

ok i see that

Answer 16

Woops I dropped the square root, my bad.

Answer 17

dw i still understood

Answer 18

\[\Large =-30\int\limits\limits\frac{1}{\sqrt{1-u^2}}\left(\frac{1}{5}du\right)\]

Answer 19

Let's pull the 1/5 out of the integral.

Answer 20

\[\Large =-6\color{orangered}{\int\limits \frac{1}{\sqrt{1-u^2}}du}\]

Answer 21

Does the orange part look familiar? :O

Answer 22

like in the first step so equals arcsinx?
so answer is -6arcsinx?

Answer 23

i mean 'u'

Answer 24

ya good catch :)
so as a final step, undo the substitution.

Answer 25

but didnt we make an error, should it be in the substitution u=25x?

Answer 26

oh sorry my bad

Answer 27

lol you catch these things so quickly XD

Answer 28

no thats right, answer being -6arcsin(5x)

Answer 29

yay team \c:/

Answer 30

Ya, the 25 isn't being squared.
We want to replace the thing being spared with u because we want that exponent on our u.

So we had to bring the 25 into the square by writing it as 5^2

Answer 31

thank you

Answer 32

i have another question if you don't ming @zepdrix

Answer 33

kk