IIT JEE , AIEEE Forum - Mathematics , Integral Calculus

ramkumar_november

here is the solution.....

I = $\int\frac{dx}{sinx+secx}$

I = $\int\frac{cosx\;\; dx}{sinxcosx+1}$

I = $\frac{1}{2}\;\;\int\frac{(cosx-sinx+cosx+sinx)\;\; dx}{sinxcosx+1}$

split into two different integrals......

$I_1\;=\;\frac{1}{2}\;\;\int\frac{(cosx-sinx)\;\; dx}{sinxcosx+1}$

$I_2\;=\;\frac{1}{2}\;\;\int\frac{(cosx+sinx)\;\; dx}{sinxcosx+1}$

I₁ can be solved by substituting sinx+cosx = t

I₂ can be solved by substituting sinx-cosx=t ......

then they become standard integrals which can be solved easily........