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Community Discussion Question: solve

Forum Index -> Integral Calculus

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15 May 2008 20:59:09 IST

Subject: solve

gokusuper (5)

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$\int$ (sec x+cosec x)^-1 dx

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23 May 2008 13:36:27 IST

Subject: Re:solve

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flowers_rsss (170)

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$\int \frac{1}{secx+cosecx}=\int \frac{sinxcosx}{sinx+cosx}=\int\frac{sin2x}{2(sinx+cosx)}$

$\int \frac{(sinx+cosx)^2-sin^2x-cos^2x}{2(sinx+cosx)}$

$= \int [(sinx+cosx)- \frac{1}{sinx+cosx}] *\frac{1}{2}$

= $\frac{1}{2} (sinx-cosx-1/\sqrt{2} log tan (x/2+\pi/8 ) )$

correct if iam wrong!

**** hey i forgot to write dx in its place jst see it

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23 May 2008 23:41:03 IST

Subject: solve

paddy.dude (1154)

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Olaaa!! Perrrfect answer.

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gr8 answer

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	Forum Index -> Integral Calculus

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