It would be better if you can post your questions here. I might be able to help, but I can't use voice channels right now.
#Need help with Calc 2 Trig Substition
winged bay
slim fog
Okay, would you like to work out a problem here then?
winged bay
I can try if it's not too advanced.
but anyway I need to see the question before deciding whether I'm able to solve.
winged bay
OK, so try $x=\sec\theta$ so that $dx=\sec\theta\tan\theta d\theta$, and we have $$\int\dfrac{\sqrt{x^2-1}}{x^4}dx=\int\dfrac{\tan\theta\cdot \sec\theta\tan\theta d\theta}{\sec^4\theta}$$
$$=\int\dfrac{\tan^2\theta\sec\theta}{\sec^4\theta}d\theta=\int\dfrac{\sin^2\theta\sec^3\theta}{\sec^4\theta}d\theta$$
$$=\int\sin^2\theta\cos\theta=\dfrac{\sin^3\theta}{3}+C=\dfrac{(\sqrt{x^2-1})^3}{3x^3}+C$$
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Kevin S
winged bay
**Kevin S**
The key is that you want to eliminate the radicals using trig identities and we have $\sec^2\theta-1=\tan^2\theta$, so $x=\sec\theta$ substitution is sensible.
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Kevin S
slim fog
I had the right identity I just subbed the wrong side of the equation I see
We want to have the x^2-1 obviously be the same side as sec^2-1 and not the lone tan
how does the sqrt(sin^2-1) become tan*sectan in the numerator of the integral, is that because of multiplying by the dx?
yeah because the derivative of sec is sectan and thats the proper sub
winged bay
how does the sqrt(sin^2-1) become tan*sectan in the numerator of the integral, i...
that is basically what happens after sub $x=\sec\theta$ cuz you also need to change dx to $\sec\theta\tan\theta d\theta$
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Kevin S
winged bay
can we try this one?
sure, use $t=4\sec\theta$ and see if you can proceed.
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Kevin S
slim fog
dx is 4sectan(dtheta) correct?
winged bay
dx is 4sectan(dtheta) correct?
yes
slim fog
apologies for the mess. and int((4sectan(theta))/(16sec^2(theta)sqrt((4(sec^2-1))^2))d(theta) is the correct sub?
maybe not sec^2 on bottom under radical, just sec?
winged bay
apologies for the mess. and int((4sectan(theta))/(16sec^2(theta)sqrt((4(sec^2-1)...
why is there a 14?, it should be $4^2=16$ and the radical get eliminated and results in $4\tan\theta$
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Kevin S
slim fog
its (4(sec-1))^2
winged bay
then it should be $(4\tan\theta)^2$
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Kevin S
winged bay
$$=\int\dfrac{4\sec\theta\tan\theta d\theta}{16\sec^2\theta\sqrt{16(sec^2\theta-1)}}$$
slim fog
then what? you get 1/4(int(sectan(theta)/4sec^2(theta)(4tan(theta))^2)d(theta) I think
and then do you multiply the bottom or do we do a trick in order to make it easier
winged bay
that square around the tangent in the denominator cancels with the square root sign.
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Kevin S
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Kevin S
winged bay
**Kevin S**
use this.
slim fog
okay so not tan^2 just tan right, other than that its correct got it
and from here...I get stuck again...do we break it up? or do they just cancel out to get 1/4sec=(1/4)cos (1/4)int((1/4)cos(theta))=(1/16)sin(theta)
winged bay
**Kevin S**
$$=\int\dfrac{4\sec\theta\tan\theta d\theta}{16\sec^2\theta 4\tan\theta}=\int\dfrac{\sec\theta d\theta}{16\sec^2\theta}=\dfrac{1}{16}\int\cos\theta d\theta$$
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Kevin S
slim fog
oops, said reciprocal of sec was sin, its cos
now integral to sin, we sub back in in terms of x to get (1/16)sin(sqrt(x^2-16)/x)+C right? 😄
so final is
(sin(sqrt(x^2-16)/16x)+C
winged bay
so final is
(sin(sqrt(x^2-16)/16x)+C
there shouldn't be any sine in the final answer. We have $$\dfrac{\sin\theta}{16}+C=\dfrac{\sqrt{t^2/16-1}/(t/4)}{16}+C=\dfrac{\sqrt{t^2-16}}{16t}+C$$
real portalBOT
Kevin S
