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#1 Posted 3:04pm 14-03-10 integration help.DOUBT[im]http://codecogs.izyba.com/gif.latex?\int_{-\pi%20/2}^{\pi%20/2}{\sqrt{cos^{2n-1}x-cos^{2n+1}x}}dx,n\epsilon%20N-[/im] ans------> [frac]4[/]2n+1[/frac]
In every loss, in every lie, in every truth that you'd deny
And each regret and each goodbye was a mistake too great to hide
And your voice was all I heard that I get what I deserve
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#2 Posted 3:20pm 14-03-10 Re: integration help.DOUBT[im]http://codecogs.izyba.com/gif.latex?\int%20\sqrt{cos^{2n%20-1}x}(\sqrt{1%20-%20cos^{2}x}dx%20=%20\int%20\sqrt{cos^{2n%20-1}x}(sinx)dx[/im]..take cosx = t..since it's a symmetric function so it can be written as [im]http://codecogs.izyba.com/gif.latex?\int_{-\pi%20/2}^{\pi%20/2}{}%20\sqrt{cos^{2n%20-1}x}(sinx)dx%20=%202\int_{0}^{\pi%20/2}{}%20\sqrt{cos^{2n%20-1}x}(sinx)dx[/im] take cosx = t [im]http://codecogs.izyba.com/gif.latex?\Rightarrow%20-\int_{1}^{0}{t^{n%20-%201/2}}dt%20=%20-\left(\frac{t^{n+1/2}}{n+1/2}%20\right)[/im] Now put the limits ..u will get the answer.. |
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