Question

need help with integration can anyone tell me the first 2 step of this question http://www.wolframalpha.com/widget/widgetPopup.jsp?p=v&id=c79bb048121bbc1d20d79c6b83ef17b5&title=Integral%20Calculator&theme=blue&i0=%28cos^4%28x%29%29%2F%28sin^3%28x%29%29%28sin^5x%2Bcos^5x%29^3%2F5&i1=x&podSelect=

Answer 1

this is all i can help, sorry i just started integrals so this is pretty hard for me to solve by my self :S

Answer 2

are you sure you entered in the correct function? is the exponent supposed to be "3/5" ?

Answer 3

yeah it is

Answer 4

3/5

Answer 5

and is the (sin^5 +cos^5)^(3/5) part supposed to be in numerator or denominator?

Answer 6

see thsi http://snag.gy/tBpM1.jpg

Answer 7

deno

Answer 8

ok good , well be careful with your parenthesis because wolfram did a completely different function

Answer 9

okay :) but do u know how to solve it

Answer 10

@hartnn -here is the question http://snag.gy/tBpM1.jpg 2 nd one

Answer 11

?????????

Answer 12

u cahnged the whole question :(

Answer 13

http://www.wolframalpha.com/input/?i=integrate+%28cos%5E4%28x%29%29%2F%28sin%5E3%28x%29%28sin%5E5x%2Bcos%5E5x%29%5E%283%2F5%29%29+dx

Answer 14

that's not the solution

Answer 15

the choices scream at me, they tell me to put u= tan x

Answer 16

i know wlfalpha can give u answer but they give compliacted answers sometimes

Answer 17

so, multiply numerator and denominator by tan^3 x

Answer 18

hmm then

Answer 19

then you need to do some simplifications

Answer 20

sorry, not by tan^3x, but by cos^3 x

Answer 21

oh

Answer 22

\(\large \dfrac{(\sin^5 x +\cos^5 x)^{3/5}}{\cos^3x} = (\dfrac{\sin^5x+\cos^5x}{\cos^5x})^{3/5} \)
did you get this ?

Answer 23

i think we have to use subtsitution t=tanx wait

Answer 24

yes, absolutely, we will use t= tan x
but for that we need a function in tan x

Answer 25

how come cos^3 x become cos^5 x when it goes inside the root

Answer 26

its not square root, its ^3/5

Answer 27

okay

Answer 28

i m getting
dt/t^6 * (t^5+1)^3/3

Answer 29

oops
dt/t^6 * (t^5+1)^3/5

Answer 30

when you push something inside square root(exponent =1/2), you push it with the exponent of 2, right ?
so, when i push something inside exponent of 3/5, i push it with the exponent of 5/3
(cos^3 x)^(5/3) = cos^5x
you got this part right ?

Answer 31

oh yeah

Answer 32

and lets not hurry
numerator is now cos x right ?

Answer 33

we need to take care of the cos x/sin^3 x part
can you express this in terms of tan x and sec^2 x ?

Answer 34

yep

Answer 35

do you get that as sec^2 x/ tan^3 x ?

Answer 36

i have a question

Answer 37

what i did wrong

Answer 38

yes ?

Answer 39

how are you getting x^6 ?

Answer 40

no i m trying to do this question since more than 1.5 hours

Answer 41

only in one stpe i m having some problem