Question

integration of (sinx.e^x-(sin x + cos x)e^(sinx+cosx))/e^2sinx-2.e^sinx+1)

Answer 1

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Answer 2

Answer:

$\frac {{ e ^ cos x }{ e^sinx - 1}} + c$

Step-by-step explanation:

$Let\quad I=∫\frac { e^{ cosx }sin\quad x-e^{ (sinx+cosx) }(sinx+cosx) }{ e^{ z\quad sinx }-e^{ sin\quad x }+1 } dx$

=∫$\frac { -e^{ cos\quad x }(e^{ sin\quad x }-1)+e^{ sin\quad x }cos\quad x }{ (e^{ sin\quad x }-1)^{ 2 } } dx$

Put $e^{cos x}^{-1} =z$ ⇒dz=e^{cos x}(-sinx)dx

and $e^{{sin x}^{-1)}=t$⇒$e^{sin x}^{-1 }cosxdx=dt$

∴I = ∫$\frac { -e^{ cos\quad x }(-sinx+e^{ sinx }sinx+e^{ x }cosx) }{ (e^{ sin\quad x }-1)^{ 2 } } dx$

∴I = ∫$\frac {tdz –zdt} {t^2}$

= $\frac { z }{ t } + c$

I = $\frac {{ e ^ cos x }{ e^sinx - 1}} + c$