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integral of 1/(sinX +secX) dx

multiply numerator and denominator by sec2x 

sec2x/(tanx+sec2x) =sec2x/(tanx+1+tan2x)

let tanx =t

=> sec2x dx =dt

and solve the eq.in the form of 't'.

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some mistake in ur suggestion,

when you multiply & divide sec2 x u will get it as sec2 x/(tan xsec x+sec3 x) dx.

but, exact answer is too long and still many of them didn't solve.!!!!!!  

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 convery in sin and cos form-

cosx/(1+sinx cosx)

 

then rationalise. it will be ((cosx/(1-sin^2x cos^2x) - (cos^2x sinx/(1-cos^2x sin^2x))

then put sinx=t in one part and cosx = t in other. solve as usual..

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This is the only way to solve this question. It cannot be solved by any other method

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Integration of 1/sin×+sec×

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I didn't find any short method other than this..... So, If anyone have please give........

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1/sinx+secx dx
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