Question

Use trig substitution
integral sqrt(x^2-1)/x^3 x=-2..-8
-2 is the lower, -8 is the upper,
i started x = sec theta

Answer 1

x^2 -1 = tan^2 theta,
dtheta = sectheta tan theta

Answer 2

i put it into my calculator and i get funny answers

Answer 3

I get -.353, but the answer according to calculator is .353

Answer 4

I have integral sqrt (tan^2 t ) / tan^3 t * sec t * tan t dt , where t = theta

Answer 5

errr
I have integral sqrt (tan^2 t ) / sec^3 t * sec t * tan t dt , where t = theta

Answer 6

, so integral tan^2 / sec ^2 dt = integral sec^2 - 1 / (sec^2

Answer 7

integral 1 - integral 1/sec^2

Answer 8

integral 1 - integral cos^2

Answer 9

\[\int\limits_{}^{}\frac{\sqrt{x^2-1}}{x^3} dx=\int\limits_{}^{}\frac{x}{x} \cdot \frac{\sqrt{x^2-1}}{x^3} dx\]
\[=\int\limits_{}^{}\frac{x \cdot \sqrt{x^2-1}}{x^4} dx\]
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Let
\[u=\sqrt{x^2-1} => u^2=x^2-1 => 2 u du =2x dx => u du =x dx\]
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\[=\int\limits_{}^{}\frac{u \cdot u}{(u^2+1)^2} du=\int\limits_{}^{}\frac{u^2}{(u^2+1)^2} du\]
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\[\frac{u^2}{(u^2+1)^2}=\frac{Au+B}{u^2+1}+\frac{Cu+D}{(u^2+1)^2}\]
\[=\frac{(Au+B)+(u^2+1)(Cu+D)}{(u^2+1)^2}\]
\[=\frac{Au+B+Cu^3+Du^2+Cu+D}{(u^2+1)^2}\]
=>
\[u^2=Au+B+Cu^3+Du^2+Cu+D=u^3(C)+u^2(D)+u(A+C)+(B+D)\]
=>
\[C=0; D=1; A=0;B=-1\]
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so we have
\[\int\limits_{}^{}\frac{u^2}{(u^2+1)^2} du=\int\limits_{}^{} (\frac{-1}{u^2+1}+\frac{1}{(u^2+1)^2}) du\]
\[=-\arctan(u)+\int\limits_{}^{}\frac{1}{(u^2+1)^2} du+C\]

Answer 10

integral 1 - integral ( 1 + cos (2x)) / 2
t - t/2 - sin t *cos t / 2

Answer 11

you should use trig substitution

Answer 12

\[\tan(\theta)=u => \sec^2(\theta) d \theta=du \]
so we have
\[\int\limits_{}^{}\frac{1}{(\tan^2(\theta)+1)^2} \sec^2(\theta) d \theta\]

Answer 13

1/2 ( t - sin t * cos t ) , where t is theta
1/2 [ arcos (-1/8) - sin (arcos(-1/8)) cos (arcos(-1/8)] - arcos (-1/2) + sin (arcos -1/2) cos (arcos-1/2) )

Answer 14

\[\int\limits_{}^{}\frac{\sec^2(\theta)}{\sec^4(\theta)} d \theta=\int\limits_{}^{}\cos^2(\theta) d \theta\]
\[\frac{1}{2}\int\limits_{}^{}(1+\cos(2\theta)) d \theta =\frac{1}{2} (\theta+\frac{1}{2}\sin(2 \theta))+C\]
\[=\frac{1}{2}(\theta+\frac{1}{2} \cdot 2 \sin(\theta) \cos(\theta))+C\]
\[=\frac{1}{2} \theta+\frac{1}{2}\sin(\theta) \cos(\theta) +C\]
\[=\frac{1}{2}\tan^{-1}(u)+\frac{1}{2} \cdot \frac{u}{\sqrt{u^2+1}} \cdot \frac{1}{\sqrt{u^2+1}}+C\]
\[=\frac{1}{2}\tan^{-1}(u)+\frac{1}{2} \cdot \frac{u}{u^2+1}+C\]
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so we have so far
\[-\arctan(u)+\frac{1}{2}\arctan(u)+\frac{u}{2(u^2+1)}+C\]
now we need to put terms of x

Answer 15

can you check over my work , and im curious what you get as a final answer

Answer 16

\[=-\arctan(\sqrt{x^2-1})+\frac{1}{2}\arctan(\sqrt{x^2-1})+\frac{\sqrt{x^2-1}}{2((x^2-1)+1)}+C\]
and i guess we could combine are like terms
\[=\frac{-1}{2}\tan^{-1}(\sqrt{x^2-1})+\frac{\sqrt{x^2-1}}{2x^2}+C\]

Answer 17

\[\int\limits_{-8}^{-2}\frac{\sqrt{x^2-1}}{x^3} dx=[-\frac{1}{2}\tan^{-1}(\sqrt{x^2-1})+\frac{\sqrt{x^2-1}}{2x^2}]_{-8}^{-2}\]
=
\[[\frac{-1}{2}\tan^{-1}(\sqrt{4-1})+\frac{\sqrt{4-1}}{8}]\]
-
\[[-\frac{1}{2}\tan^{-1}(\sqrt{64-1})+\frac{\sqrt{64-1}}{128}\]

Answer 18

my calculator gives me an answer of .3536 , but when i do it out i get -.3536 , im not sure what i did wrong

Answer 19

=
\[\frac{-1}{2}\tan^{-1}(\sqrt{3})+\frac{\sqrt{3}}{8}+\frac{1}{2}\tan^{-1}(\sqrt{63})-\frac{\sqrt{63}}{128}\]

Answer 20

yes i get .3536

Answer 21

did you do F(upper limit)-F(lower limit)

Answer 22

oh wait the upper limit is -8?

Answer 23

i thought it was the lower limit

Answer 24

it would -.3536 then

Answer 25

yes