1. anonymous

This one is a bit more involved. It is again separable, so$\frac{dy}{y.\ln(y)}=\frac{dx}{\sin(x)}$ The left-hand side can be integrated using the substitution u=ln(y), for then du=(1/y)dy and we can re-write the left-hand side as$\frac{du}{u} \rightarrow \int\limits_{}{}\frac{du}{u}=\ln(u)+c_1$Substituting u=ln(y) back in, we get$\ln(\ln(y))+c_1$as solution to the left-hand side. I'll post the right-hand side next...I have to break it up since this site is awkward to use when typing large pieces.

2. anonymous

You have to use a t-substitution on the right-hand side. I was going to go into setting it up, but this site is difficult to use. I'm going to assume you have heard of the t-substitution (if not, it's easy to find out information). From this substitution technique, we have two results to use:$\sin(x)=\frac{2t}{1+t^2}$ and $dx=\frac{2dt}{1+t^2}$Plugging these into dx/sin(x) we have$\int_{}{}\frac{dx}{\sin(x)} \rightarrow \int\limits_{}{}\frac{1+t^2}{2t}.\frac{2dt}{1+t^2}=\int\limits_{}{}\frac{dt}{t}=\ln(t)+c_2$Now, from the definition of t-substitution, $t=\tan(\frac{x}{2})$so upon substituting, the right-hand side becomes$\ln(\tan(\frac{x}{2}))+c_2$

3. anonymous

nice i see)0

4. anonymous

5. anonymous

Okay, now we bring the two answers together,$\ln(\ln(y))+c_1=\ln(\tan(\frac{x}{2}))+c_2$Collecting the constants on the RHS and exponentiating, we get,$e^{\ln(\ln(y))}=e^{\ln(\tan(\frac{x}{2}))+(c_2-c_1)}$Which simplifies to$\ln(y)=e^{(c_2-c_1)}e^{\ln(\tan(\frac{x}{2}))}=A{\tan(\frac{x}{2})}$where A is just e^(c_2-c_1), a constant. Exponentiating again isolates y,$e^{\ln(y)}=e^{Atan(\frac{x}{2})}=e^Ae^{\tan(\frac{x}{2})}=Be^{\tan(\frac{x}{2})}$where again, B is just the collected constant, e^A. So you're answer is $y=Be^{\tan(\frac{x}{2})}$

6. anonymous

Grammatical error - *your* answer, not, *you're*.

7. anonymous

Remember to check everything. You can do that by differentiating the final answer - you should be able to get back to the original equation. I haven't checked anything I've sent, so please do that. Good luck. Bed time...

8. anonymous

PS - I'm glad you enjoyed the responses!

9. anonymous

wow!!))thanx very much, sir,, i have really enjoyed your responses ,