Question

I need help with trigonometric substitions

Intergral of (sqrt(y^2-49)/y)dy
I know that y will be equal to 7sec theta
My solutions manual has a funny answer that I can't get to. I need help working it out, thanks!

Answer 1

\[\int\limits (\sqrt{y^2-49})/y) dy\]

Answer 2

\[\int\limits_{}^{}\frac{\sqrt{49\tan^2{\theta}-49}}{7\tan \theta} 7 \sec^2{\theta} d \theta\]
\[\frac{\sqrt{49}*7}{7}\int\limits_{}^{}\frac{\sqrt{\tan^2{\theta}-1}}{\tan \theta} \sec^2{\theta} d \theta\]

Answer 3

so we can conclude that tan was the wrong substituition

Answer 4

My solutions manual has this:
y=7sec\[\theta\]
dy=7sec\[\theta\]tan\[\theta\]d\[\theta\]

Answer 5

oh darn...i wonder why it spaces it like that

Answer 6

y=7sec theta
dy=7sec theta tan theta
and therefore \[\sqrt{(y^2-49)}\] is equal to 7tan theta

Answer 7

but i have no clue how they got 7 tan theta for the \[\sqrt{y^2-49}\]

Answer 8

oops thats wrong i cant think today i sneezed so much

Answer 9

Haha, I was all confused, lol!

Answer 10

It's all good tho

Answer 11

ok im going to write this on paper and post it k? give me a few

Answer 12

Oh thanks!

Answer 13

\[7\int\limits_{}^{}\tan^2{\theta} d \theta=7\int\limits_{}^{}(\sec^2{\theta}-1) d \theta=7\int\limits_{}^{}\sec^2{\theta} d \theta-7\int\limits_{}^{}1 d \theta\]
=\[7\tan \theta-7 \theta+C\]
we need to put this in terms of x

Answer 14

or i mean in terms of y lol
do you see our traingle?
\[\tan \theta=\frac{opposite}{adjacent}=\frac{\sqrt{y^2-49}}{7}\]

Answer 15

\[\sec{\theta}=\frac{y}{7}\]
so
\[\theta=\sec^{-1}(\frac{y}{7})\]

Answer 16

Yes! Thank you so much! Duh, I can't believe I didn't see that!

Answer 17

\[7*\frac{\sqrt{y^2-49}}{7}-7*\sec^{-1}(\frac{y}{7})+C\]

Answer 18

those 7 cancel in the first fraction

Answer 19

but u knew that

Answer 20

it always helps me to draw a traingle for these type of problems

Answer 21

thats what i should had done at the very beginining

Answer 22

Yea, I need to draw the triangle first thing, I always do it last, hehe

Answer 23

drawing those three triangles helps me to determine which substituition to use
also it helps when you need to put everything back in terms of whatever it was

Answer 24

try another one and if you can't get it post it

Answer 25

Awesome, thanks!