Integration problem calculus

Re: Integration problem calculus

Quote:

Originally Posted by

**danny88**

Hints: you have $\displaystyle \arctan(x)+\arctan(y)=\arctan(z)$

So $\displaystyle \tan(a+b)=z$

Re: Integration problem calculus

I really don't get the next step

Re: Integration problem calculus

Quote:

Originally Posted by

**danny88** I really don't get the next step

What is $\displaystyle \tan(a+b)=$$

Re: Integration problem calculus

Re: Integration problem calculus

Quote:

Originally Posted by

**danny88** sin(A+B)/cos(A+B)

That does you no good!

$\displaystyle \tan(a+b)=\frac{\tan(a)+\tan(b)}{1-\tan(a)\tan(b)}$.

Clearly $\displaystyle a=\arctan(x)$.

If you are still confused, then it is time to leave this site and seek live instruction from your instructor.

Re: Integration problem calculus