If y=y(x) and 2+sinxy+1dydx=-cosx,y(0)=1, then yπ2 equals

−∫dyy+1=∫cosx2+sinxdx Put t=2+sinxSince∫f'(x)f(x)dx=log|f(x)|+c⇒-ln|y+1|=ln|t|+lnk ⇒(sinx+2)(y+1)=k1 ⇒2(1+1)=k1⇒k1=4 ⇒(3)(y+1)=4⇒y=13