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Posted January 27th, 2010 by alpha
in

$ \int\sqrt{\frac{cos^3x}{sin^{11}x}}.dx $

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My Pocedure:

On January 27th, 2010 alpha says:

$ \int \frac{cos^{3/2}x dx }{sin^{11/2}x} $
=$ \int\frac{cos^{3/2}x dx}{sin^{5/2}x . sin^{6/2}x} $
Let $ sin^{5/2}x=t $

Here is your mistake!

$$dt=\frac{5}{2}\sin^{\frac{3}{2}}\cos x dx$$

then, $ cos^{3/2}x dx=\frac{2dt}{5} $
$ =\frac{2}{5}\int t^{-11/5} dt $
And finally I got,
$ \frac{-22}{5}. sin^{-8}x+C $
But my answer does not match with the answer given in the book.
The answer(given in the book) is$ \frac{-2}{5}cot^{5/2}x-\frac{2}{9}cot^{9/2}x+c $

-------------------x----------------------------
``Old theorems never die; they turn into definitions.''
----E. Hewitt

''ALPHA" =)

Thanks for finding my mistake.Thanks a lot.

On January 27th, 2010 alpha says:

-------------------x----------------------------
``Old theorems never die; they turn into definitions.''
----E. Hewitt

''ALPHA" =)

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