Question

integral cosx/1+sin^2x dx

Answer 1

Answer :

Let us take  

    sinx = z

    ⇒ cosx dx = dz

Then,  

    $\bold{\int \dfrac{cosx}{1+sin^{2}x}dx}$

    $\bold{= \int \dfrac{dz}{1+z^{2}}}$

    $\bold{= tan^{-1}z}$

    $\bold{= tan^{-1}(sinx)}$

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