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    How to integral 1/(1-sinx^2)^3/2? :confused::confused::confused:
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    edit: oops, do you mean sin(x²) or (sin(x))² ?

    2nd edit: why has this been negged? the difference between the two i mentioned above is pretty big: one can be done, the other cant...
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    i mean sin(x²)
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    (Original post by jackyc007)
    i mean sin(x²)
    In which case, this integral cannot be expressed in terms of elementary functions.

    If, however, you meant this:

    Integrate \displaystyle\int \dfrac{1}{(1-\sin ^2x)^{\frac{3}{2}}}dx

    Then note that 1-\sin ^2x =\cos ^2x and \dfrac{1}{cos ^n x} = \sec ^n x.
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    See the integral of secant cubed here on Wikipedia.
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    check out wolframalpha.com.
    You can integrate functions using:
    integrate[1/(1-sin(x)^2)^(3/2)]

    what (x+1)' does is obvious I guess.

    I have been using this tool when I had problems solving a question and it has been a great help. You can also list the steps. If ie Mathematica is too expensive, wolframalpha is a great alternative.
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    (Original post by Chr0n)
    check out wolframalpha.com.
    You can integrate functions using:
    integrate[1/(1-sin(x)^2)^(3/2)]

    what (x+1)' does is obvious I guess.

    I have been using this tool when I had problems solving a question and it has been a great help. You can also list the steps. If ie Mathematica is too expensive, wolframalpha is a great alternative.
    I think it should be pointed out that using Wolfram for anything but checking your answers isn't going to be beneficial for any exam you must take on the subject! It may well do your homework for you but you won't benefit from just looking at the answer.
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     \displaystyle \int\frac{1}{\cos^3{x}}\;{dx} = \int\frac{\cos{x}}{\cos^4{x}}\;{  dx} = \int\frac{\cos{x}}{\left(1-\sin^2{x}\right)^2}\;{dx}, let  t = \sin {x} then:
    \displaystyle \int\frac{1}{\cos^3{x}}\;{dx} = \int\frac{1}{(1-t^2)^2}\;{dt} = \int\frac{1}{(1-t)^2(1+t)^2}\;{dt}. Observe that:

    \displaystyle \frac{1}{(1-t)^2(1+t)^2} = \frac{(1+t)^2-(1-t)^2}{4t(1-t)^2(1+t)^2} = \frac{1}{4t(1-t)^2}-\frac{1}{4t(1+t)^2}, where we can write:
    \displaystyle \frac{1}{4t(1-t)^2} = \frac{(1-t)+t}{4t(1-t)^2} = \frac{(1-t)+t}{4t(1-t)}+\frac{1}{4(1-t)^2} = \frac{1}{4t}+\frac{1}{4(1-t)}+\frac{1}{4(1-t)^2} and
    \displaystyle \frac{1}{4t(1+t)^2} = \frac{(1+t)-t}{4t(1+t)^2} = \frac{(1+t)-t}{4t(1+t)}-\frac{1}{4(1+t)^2} = \frac{1}{4t}-\frac{1}{4(1+t)}-\frac{1}{4(1+t)^2}, thus:

    \displaystyle \begin{aligned} & \int\frac{1}{(1-t)^2(1+t)^2}\;{dt}  = \int\frac{1}{4(1-t)^2}\;{dt}+\int\frac{1}{4(1+t)^  2}\;{dt}+\int\frac{1}{4(1-t)}\;{dt}+\int\frac{1}{4(1+t)}\;  {dt} \end{aligned}
    \displaystyle = \frac{1}{4(1-t)}-\frac{1}{4(1+t)}-\frac{1}{4}\ln\left(1-t\right)+\frac{1}{4}\ln\left(1+t  \right)  = \frac{t}{2(1-t^2)}+\frac{1}{4}\ln\left(\frac{  1+t}{1-t}\right), therefore:

    \displaystyle \int\frac{1}{\cos^3{x}}\;{dx} = \frac{\sin{x}}{2(1-\sin^2{x})}+\frac{1}{4}\ln\left(  \frac{1+\sin{x}}{1-\sin{x}}\right). Beat that, Wiki and Wolfram! Looks pleasing too.
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    Wow @ piecewise
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    Simpler:

    Let I = \int \sec^3 x\, dx. IBP with dv = sec^2 x (so v = tan x) to get I = \tan x  \sec x  - \int \tan x \sec x \tan x \, dx. Write J for the RH integral, so that

    J  =  \int \tan^2 x \sec x \, dx = \int (\sec^2 x -1)\sec x \, dx = \int \sec^3 x - \int \sec x \, dx = I - \int \sec x\,dx.

    But I = \tan x \sec x - J, so I = \tan x \sec x - I + \int \sec x\,dx and so 2I = \tan x \sec x  + \ln |\sec x + \tan x|.
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    (Original post by Piecewise)
     \displaystyle \int\frac{1}{\cos^3{x}}\;{dx} = \int\frac{\cos{x}}{\cos^4{x}}\;{  dx} = \int\frac{\cos{x}}{\left(1-\sin^2{x}\right)^2}\;{dx}, let  t = \sin {x} then:
    \displaystyle \int\frac{1}{\cos^3{x}}\;{dx} = \int\frac{1}{(1-t^2)^2}\;{dx} = \int\frac{1}{(1-t)^2(1+t)^2}\;{dt}. Observe that:

    \displaystyle \frac{1}{(1-t)^2(1+t)^2} = \frac{(1+t)^2+(1-t)^2}{4t(1-t)^2(1+t)^2} = \frac{1}{4t(1-t)^2}-\frac{1}{4t(1+t)^2}, where we can write:
    \displaystyle \frac{1}{4t(1-t)^2} = \frac{(1-t)+t}{4t(1-t)^2} = \frac{(1-t)+t}{4t(1-t)}+\frac{1}{4(1-t)^2} = \frac{1}{4t}+\frac{1}{4(1-t)}+\frac{1}{4(1-t)^2} and
    \displaystyle \frac{1}{4t(1+t)^2} = \frac{(1+t)-t}{4t(1+t)^2} = \frac{(1+t)-t}{4t(1+t)}-\frac{1}{4(1+t)^2} = \frac{1}{4t}-\frac{1}{4(1+t)}-\frac{1}{4(1+t)^2}, thus:

    \displaystyle \begin{aligned} & \int\frac{1}{(1-t)^2(1+t)^2}\;{dt}  = \int\frac{1}{4(1-t)^2}\;{dt}+\int\frac{1}{4(1+t)^  2}\;{dt}+\int\frac{1}{4(1-t)}\;{dt}+\int\frac{1}{4(1+t)}\;  {dt} \end{aligned}
    \displaystyle = \frac{1}{4(1-t)}-\frac{1}{4(1+t)}-\frac{1}{4}\ln\left(1-t\right)+\frac{1}{4}\ln\left(1+t  \right)  = \frac{t}{2(1-t^2)}+\frac{1}{4}\ln\left(\frac{  1+t}{1-t}\right), therefore:

    \displaystyle \int\frac{1}{\cos^3{x}}\;{dx} = \frac{\sin{x}}{2(1-\sin^2{x})}+\frac{1}{4}\ln\left(  \frac{1+\sin{x}}{1-\sin{x}}\right). Beat that, Wiki and Wolfram! Looks pleasing too.
    You have successfully found the long way of doing this integral. :p:
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    (Original post by Farhan.Hanif93)
    You have successfully found the long way of doing this integral. :p:
    Haha, the fact that we have to use the integral of secx puts me off from integration by parts.
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    (Original post by Piecewise)
    Haha, the fact that we have to use the integral of secx puts me off from integration by parts.
    Well yeah, the integral of sec x is nasty. But it's also a standard result. Much of the work you've done has been effectively finding the same integral from first principles.

    (Or to put it another way, I cheated. But in an examiner acceptable fashion).
 
 
 
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