anonymous
  • anonymous
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
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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myininaya
  • myininaya
\[\int\limits_{}^{}\sqrt{y^2-49}dy\] let \[\frac{y}{7}=\tan(\theta)\] \[\frac{dy}{7}=\sec^2\theta d \theta\] \[\int\limits_{}^{}\sqrt{(7\tan \theta)^2-49}7\sec^2\theta d \theta\]
anonymous
  • anonymous
isnt it better to use 7 sin theta as a sunstitution
myininaya
  • myininaya
\[7\int\limits_{}^{}\sqrt{49}\sqrt{\tan^2\theta-1}\sec^2(\theta)d \theta\]

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myininaya
  • myininaya
\[7*7\int\limits_{}^{}\sec \theta*\sec^2(\theta) d \theta\]
myininaya
  • myininaya
\[49\int\limits_{}^{}\sec^3(\theta) d \theta\]
myininaya
  • myininaya
oops i made the wrong substitution
anonymous
  • anonymous
yah lol
myininaya
  • myininaya
tan^2x+1=sec^2x not tan^2x-1=sec^2x
anonymous
  • anonymous
could you do it using sin theta as the substitution if it not too much trouble
anonymous
  • anonymous
this is the part im stuck if it helps
myininaya
  • myininaya
you could also use sintheta=y/7
anonymous
  • anonymous
i got 49 cos^2 theta
myininaya
  • myininaya
i mean costhetha=y/7
anonymous
  • anonymous
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
myininaya
  • myininaya
\[\sin \theta=\frac{y}{7}\] \[\cos(\theta)d \theta=\frac{dy}{7}\] \[(\sin \theta)^2=(\frac{y}{7})^2\] \[49\sin^2(\theta)=y^2\] \[\int\limits_{}^{}\sqrt{49\sin^2(\theta)-49}*7\cos(\theta)d \theta\] \[7\int\limits_{}^{}\sqrt{49}*\sqrt{\sin^2(\theta)-1}*\cos(\theta)d \theta=49\int\limits_{}^{}\cos^2\theta d \theta\]
myininaya
  • myininaya
remember cos^2(theta)=1/2*(1+cos(2theta))
myininaya
  • myininaya
\[49*\frac{1}{2}\int\limits_{}^{}(1+\cos(2\theta)) d \theta=\frac{49}{2}*(\theta+\frac{1}{2}\sin (2\theta))+C\]
myininaya
  • myininaya
but remember sin(2theta)=2sin(theta)cos(theta)
myininaya
  • myininaya
\[\frac{49}{2} \theta+\frac{49}{2} \sin(\theta)\cos(\theta)+C\]
myininaya
  • myininaya
we need this in terms of x though we let sintheta=y/7 so costhetha=sqrt(49-y^2)/7
myininaya
  • myininaya
also since sintheta=y/7 then theta=arcsin(y/7)
myininaya
  • myininaya
meant in terms of y lol
anonymous
  • anonymous
hey i forgot to give you the limits bc i am having trouble plugging it its 0 to 7
myininaya
  • myininaya
\[\frac{49}{2}\sin^{-1} (y)+\frac{49}{2}\frac{y}{7}\frac{\sqrt{49-y^2}}{7}+C\]
myininaya
  • myininaya
\[\frac{49}{2}\sin^{-1} (y)+\frac{y}{2}\sqrt{49-y^2}+C \]
myininaya
  • myininaya
you are having trouble plugging in?
myininaya
  • myininaya
wherever you see a y plug in your upper limit then minus wherever you see a y plug in your lower limit
myininaya
  • myininaya
\[(\frac{49}{2}\sin^{-1}(7)+\frac{7}{2}\sqrt{49-7^2})-(\frac{49}{2}\sin^{-1}(0)+\frac{0}{2}\sqrt{49-0^2})\]
myininaya
  • myininaya
\[\frac{49}{2}\sin^{-1}(7)\]
anonymous
  • anonymous
ty :)

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