Answered

How I can solve this integral ∫ dx / akar 25 - 16x^2 ? I really need your help, thank you.



Answer :

jacoblclinton
t ∫ 1/√(1 - x²) dx = arcsin(x) 

 ∫ 1/√(1 - x²/a) dx = √a arcsin(x/√a) 

So pull a factor of 5 out of the square root to get 
1/5 ∫ 1/√(1 - 16 x²/25) dx 
= 1/4 arcsin(4 x/5)
[tex]\int\frac{dx}{25-16x^2}\\\\\frac{1}{25-16x^2}=\frac{1}{5^2-(4x)^2}=\frac{1}{(5-4x)(5+4x)}=\frac{A}{5-4x}+\frac{B}{5+4x}=\frac{A(5+4x)+B(5-4x)}{(5-4x)(5+4x)}\\\\=\frac{5A+4Ax+5B-4Bx}{25-16x^2}=\frac{(4A-4B)x+(5A+5B)}{25-16x^2}\Rightarrow\frac{1}{25-16x^2}=\frac{(4A-4B)x+(5A+5B)}{25-16x^2}\\\Updownarrow\\4A-4B=0\ and\ 5A+5B=1\\4A=4B\ and\ 5A+5B=1\\A=B\ and\ 5B+5B=1\\A=B\ and\ 10B=1\\A=B\ and\ B=0.1\\A=0.1\ and\ B=0.1[/tex]

[tex]\int\frac{dx}{25-16x^2}=\int\left(\frac{0.1}{5-4x}+\frac{0.1}{5+4x}\right)dx\\\\=-\frac{1}{4}\times0.1\times log(5-4x)+\frac{1}{4}\times0.1\times log(5+4x)+C\\\\=\frac{log(5+4x)-log(4-5x)}{40}+C[/tex]