Question

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

Answer 1

`\int(dx)/(25-16x^2)\\\\(1)/(25-16x^2)=(1)/(5^2-(4x)^2)=(1)/((5-4x)(5+4x))=(A)/(5-4x)+(B)/(5+4x)=(A(5+4x)+B(5-4x))/((5-4x)(5+4x))\\\\=(5A+4Ax+5B-4Bx)/(25-16x^2)=((4A-4B)x+(5A+5B))/(25-16x^2)\Rightarrow(1)/(25-16x^2)=((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`


`\int(dx)/(25-16x^2)=\int\left((0.1)/(5-4x)+(0.1)/(5+4x)\right)dx\\\\=-(1)/(4)*0.1* log(5-4x)+(1)/(4)*0.1* log(5+4x)+C\\\\=(log(5+4x)-log(4-5x))/(40)+C`

Answer 2

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)