Question

Heres a fun one (; I would like to solve it with ust trig sub and no partial fractions.
Integral (x^2)/(25+x^2)^2

Answer 1

copy pasted answer. but i realised this is much easier using regular sub rather than trig sub

Answer 2

do you absolutely want trig sub?

Answer 3

Any help would be appreciated. :) But trig sub would be appreciated to see-just how I would work through it. Thank you!

Answer 4

\[\frac{ x^2 }{ (25 + x^2)^2 } = \left( \frac{ x }{ 25 + x^2 } \right)^2\]
\[x = 5 \tan \theta\]\[dx = 5 \sec^2 \theta d \theta\]

\[\int\limits_{}^{}\left( \frac{ x }{ 25 + x^2 } \right)^2 dx = \int\limits_{}^{}\left( \frac{ 5 \tan \theta }{ 25 + 25\tan^2 \theta } \right)^2 dx = \int\limits_{}^{}\left( \frac{ 5 \tan \theta }{ 25(1 + \tan^2 \theta) } \right)^2 sec^2 \theta d \theta \]\[\int\limits\limits_{}^{}\left( \frac{ \tan \theta }{ 5\sec^2\theta } \right)^2 \sec^2 \theta d \theta = \frac{ 1 }{ 25 }\int\limits_{}^{}\frac{ \tan^2 \theta }{ \sec^2\theta } d \theta = \frac{ 1 }{ 25 }\int\limits_{}^{}\frac{ \sec^2 \theta - 1 }{ \sec^2 \theta }d \theta\]\[\frac{ 1 }{ 25 }\int\limits_{}^{}\left( 1 - \cos^2 \theta \right) d \theta = \frac{ 1 }{ 25 }\int\limits_{}^{}\sin^2 \theta d \theta = \frac{ 1 }{ 25 }\int\limits_{}^{}\frac{ 1 - \cos(2 \theta) }{ 2 } d \theta\]\[\frac{ 1 }{ 50 }\int\limits_{}^{}(1 - \cos(2 \theta)) d \theta\]

Answer 5

let me know if i should finish it

Answer 6

Spot on, great job @Euler271 .

Answer 7

And I agree, normal substitution would have used less steps incoming to a conclusion.

Answer 8

thanks ^_^

Answer 9

made a small mistake.

i forgot to implement the 5 in\[dx = 5 sec^2 \theta d \theta\]
5/50 = 1/10, so the last line should read\[\frac{ 1 }{ 10 }\int\limits_{}^{}[1 - \cos(2 \theta)]d \theta\]

Answer 10

Thanks Euler I will keep working on it and let you know if I hav any trouble should be ok from here....thanks for your help!

Answer 11

glad i could help ^_^

don't hesitate to ask questions if any of it is unclear. i skipped visual steps and used a few trig identities

Answer 12

\[\int\limits_{}^{}\left( \frac{ x }{ 25 + x^2 } \right)^2 dx = \int\limits_{}^{}\left( \frac{ 5 \tan \theta }{ 25 + 25\tan^2 \theta } \right)^2 dx = \int\limits_{}^{}\left( \frac{ 5 \tan \theta }{ 25(1 + \tan^2 \theta) } \right)^2 5sec^2 \theta d \theta\]\[\int\limits\limits_{}^{}\left( \frac{ \tan \theta }{ 5\sec^2\theta } \right)^2 5sec^2 \theta d \theta = \frac{ 1 }{ 25 }\cdot5\int\limits_{}^{}\frac{ \tan^2 \theta }{ \sec^2\theta } d \theta = \frac{ 1 }{ 5 }\int\limits_{}^{}\frac{ \sec^2 \theta - 1 }{ \sec^2 \theta }d \theta\]\[\frac{ 1 }{ 5 }\int\limits_{}^{}\left( 1 - \cos^2 \theta \right) d \theta = \frac{ 1 }{ 5 }\int\limits_{}^{}\sin^2 \theta d \theta = \frac{ 1 }{ 5 }\int\limits_{}^{}\frac{ 1 - \cos(2 \theta) }{ 2 } d \theta\]\[\frac{ 1 }{ 10 }\int\limits_{}^{}(1 - \cos(2 \theta)) d \theta\]

Answer 13

Just helping you out ^___^

Answer 14

How do you get from \[Sec^2\theta-1/\sec^2\theta \to 1-\cos^2\Theta?\]

Answer 15

O I see sorry (;