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sat_chen
Group Title
Can someone explain me this trigonometric substitution
 one year ago
 one year ago
sat_chen Group Title
Can someone explain me this trigonometric substitution
 one year ago
 one year ago

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sat_chen Group TitleBest ResponseYou've already chosen the best response.0
http://www.wolframalpha.com/input/?i=integral+of+%281%2F%28x%5E3*sqrt%28x%5E225%29%29%29dx
 one year ago

sat_chen Group TitleBest ResponseYou've already chosen the best response.0
so what i am confused is on how someone went from
 one year ago

sat_chen Group TitleBest ResponseYou've already chosen the best response.0
dw:1359567855994:dw
 one year ago

sat_chen Group TitleBest ResponseYou've already chosen the best response.0
to the solution
 one year ago

sat_chen Group TitleBest ResponseYou've already chosen the best response.0
i think you have to log in to wolfram before you can see the steps
 one year ago

sat_chen Group TitleBest ResponseYou've already chosen the best response.0
we get sin taking the integral of cos?
 one year ago

shubhamsrg Group TitleBest ResponseYou've already chosen the best response.0
Yes I can't see the steps.
 one year ago

shubhamsrg Group TitleBest ResponseYou've already chosen the best response.0
Was the substitution x = 5 siny ?
 one year ago

sat_chen Group TitleBest ResponseYou've already chosen the best response.0
yea you have to log in for that i think and it would take me a year to type the solutions here XD
 one year ago

sat_chen Group TitleBest ResponseYou've already chosen the best response.0
so someone with wolfram account tell me how they went from that step to the solution
 one year ago

shubhamsrg Group TitleBest ResponseYou've already chosen the best response.0
Leme try anyways, let x= 5secy => dx= 5 secy tany dy so we have inside integral 5 secy tany dy/( 125 sec^3 y ( 25 + 25sec^2 y)^1/2) => 5sec y tany dy/( 625 sec^3 y (tan^2 y)^1/2 ) => 1/125 ( dy/sec^2y)) or simply cos^2y dy Am I right so far ?
 one year ago

sat_chen Group TitleBest ResponseYou've already chosen the best response.0
i got a better way for you to see the solutions this is part 2 of the solution
 one year ago

sat_chen Group TitleBest ResponseYou've already chosen the best response.0
theres all the steps to the solution can you take me from that 0/250 + sin20/500 + c to the solution
 one year ago

sat_chen Group TitleBest ResponseYou've already chosen the best response.0
and yes you are right with your steps
 one year ago

sat_chen Group TitleBest ResponseYou've already chosen the best response.0
so from the integral of (1/125)(cos^20d0) you do the half angle formula to get integral of (1/125)(1+cos20)/2dp then you split and get 0/250 + sin2(theta)/500 + c so my question if how you get from here to the answer
 one year ago

sat_chen Group TitleBest ResponseYou've already chosen the best response.0
dw:1359570048133:dw
 one year ago

sat_chen Group TitleBest ResponseYou've already chosen the best response.0
i get the second part where it is the inverse of sec1 (x/5) but i dont get where they got the 5sqrt(x^2) stuff from
 one year ago

sat_chen Group TitleBest ResponseYou've already chosen the best response.0
dw:1359570931896:dw the stuff i dont get is in that box in case you are still confused on what i am confused about
 one year ago

phi Group TitleBest ResponseYou've already chosen the best response.0
you are asking about \[ \frac{1}{2} \sin(2 u)\] with \[ u= \sec^{1}\left(\frac{x}{5}\right) \] In terms of cosine, rather than secant, we can say \[ u= \cos^{1}\left(\frac{5}{x}\right) \] If we take the hypotenuse to be x, in a triangle with side 5 dw:1359572361980:dw the other side is \( \sqrt{x^225} \) Using these sides, we have \[ \cos u = \frac{5}{x} \] and \[ \sin u = \frac{\sqrt{x^225}}{x} \] sin(2u) = 2 sin(u) cos(u) sub in the above definitions wolfram then multiplied by sqrt(x^2)/x (why?) to get your boxed expression
 one year ago
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