AboutAlon Mandes Expertise Kind of questions I can answer : Limits, Derivatives, Integration, Implicit functions, continuousity, differentiation ,Extremum problems, Lagrange multipliers, Gradients, Surface integrals, Multi variables functions ,Multi variables Integrals,Complex variables ,Complex functions, Curves, Trajectory integrals & Vector analyse,Divergence,Rotor & word problems.
Kind of question I can't answer : Economics,Combinatorics,infinite series & convergence ,Statistics & Probabilities .
Experience 1. I'm a team member of mathnerds (math site for answering questions)
2. I'm a team member in the Student's Union of the Technion, helping
students who have problems in mathematics.
3. 2 years of experience as a math teacher in college.
4. I give free homework help for high school students in
Mathematics & Physics.
5. I teach part time in collage the subjects : "Digital Signal Processing" , "Random Signals & Noise" ,
"Complex Functions".
Question pleez help me to integrate 1/1 +cos x.tnx
Answer OK, if I understood right , The Integrand function is :
1/[1+cos(x)tang(x)]=1/[1+sin(x)], so our integral becomes :
∫dx/[1+sinx]. Let set t=1+sinx then dt=cosxdx=sqrt[1-sin^2(x)]dx.
So dx=dt/sqrt(1-t^2) . Now we have to solve ∫dt/t[sqrt(1-t^2)].
Well,this integral need some more work :
I can claim that :
1/t[sqrt(1-t^2)] = sqrt(1-t^2)/t + t/sqrt(1-t^2) (chick it please !)
That means, our integral now becomes :
∫t/sqrt(1-t^2) + ∫sqrt(1-t^2)/t . OK.
The first integral is easy & it is -sqrt(1-t^2) .(chick it!)
The second part is the hard one. But, we will use the same method
as before : let's set u=sqrt(1-t^2) ,it means du=-t/sqrt(1-t^2)dt
--> dt=-dusqrt(1-t^2)/t
Hence, our integral is ∫-u^2/(1-u^2). (please confirm it !)
This integral is much easier & can be solved by using the trick
-u^2=1-u^2+1, I will leave it to you as an exercise & only give
you the final answer : ∫-u^2/(1-u^2)=u+0.5Ln(u-1)-0.5Ln(u+1). &
in terms of t is :
sqrt(1-t^2)+0.5Ln(sqrt(1-t^2)-1)-0.5Ln(sqrt(1-t^2)+1)
All right, so now what do we have so far ?
∫dx/[1+sinx]=∫dt/t[sqrt(1-t^2)]=∫t/sqrt(1-t^2)+∫sqrt(1-t^2)/t
=0.5Ln(sqrt(1-t^2)-1)-0.5Ln(sqrt(1-t^2)+1). Ler's put t=1+sinx :
