Question

hey can someone please help me find the integral of sqrt of y^2-49. i know i have to use trig substitution but i need help with actually solving it

Answer 1

\[\int\limits_{}^{}\sqrt{y^2-49}dy\]
let \[\frac{y}{7}=\tan(\theta)\]
\[\frac{dy}{7}=\sec^2\theta d \theta\]
\[\int\limits_{}^{}\sqrt{(7\tan \theta)^2-49}7\sec^2\theta d \theta\]

Answer 2

isnt it better to use 7 sin theta as a sunstitution

Answer 3

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

Answer 4

\[7*7\int\limits_{}^{}\sec \theta*\sec^2(\theta) d \theta\]

Answer 5

\[49\int\limits_{}^{}\sec^3(\theta) d \theta\]

Answer 6

oops i made the wrong substitution

Answer 7

yah lol

Answer 8

tan^2x+1=sec^2x
not
tan^2x-1=sec^2x

Answer 9

could you do it using sin theta as the substitution if it not too much trouble

Answer 10

this is the part im stuck if it helps

Answer 11

you could also use sintheta=y/7

Answer 12

i got 49 cos^2 theta

Answer 13

i mean costhetha=y/7

Answer 14

then i knew i i got the integral of cos^2 theta which came out to ve 1/2 theta -1/4 cos 2 theta

Answer 15

\[\sin \theta=\frac{y}{7}\]
\[\cos(\theta)d \theta=\frac{dy}{7}\]
\[(\sin \theta)^2=(\frac{y}{7})^2\]
\[49\sin^2(\theta)=y^2\]
\[\int\limits_{}^{}\sqrt{49\sin^2(\theta)-49}*7\cos(\theta)d \theta\]
\[7\int\limits_{}^{}\sqrt{49}*\sqrt{\sin^2(\theta)-1}*\cos(\theta)d \theta=49\int\limits_{}^{}\cos^2\theta d \theta\]

Answer 16

remember cos^2(theta)=1/2*(1+cos(2theta))

Answer 17

\[49*\frac{1}{2}\int\limits_{}^{}(1+\cos(2\theta)) d \theta=\frac{49}{2}*(\theta+\frac{1}{2}\sin (2\theta))+C\]

Answer 18

but remember sin(2theta)=2sin(theta)cos(theta)

Answer 19

\[\frac{49}{2} \theta+\frac{49}{2} \sin(\theta)\cos(\theta)+C\]

Answer 20

we need this in terms of x though
we let sintheta=y/7
so costhetha=sqrt(49-y^2)/7

Answer 21

also since sintheta=y/7
then theta=arcsin(y/7)

Answer 22

meant in terms of y lol

Answer 23

hey i forgot to give you the limits bc i am having trouble plugging it its 0 to 7

Answer 24

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

Answer 25

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

Answer 26

you are having trouble plugging in?

Answer 27

wherever you see a y plug in your upper limit
then minus
wherever you see a y plug in your lower limit

Answer 28

\[(\frac{49}{2}\sin^{-1}(7)+\frac{7}{2}\sqrt{49-7^2})-(\frac{49}{2}\sin^{-1}(0)+\frac{0}{2}\sqrt{49-0^2})\]

Answer 29

\[\frac{49}{2}\sin^{-1}(7)\]

Answer 30

ty :)