Question

integration of √16-25x^2*dx​

Answer 1

Step-by-step explanation:

This is the answer..

HOPE THIS WAS HELPFUL!!

Answer 2

Answer:

The integration of $\sqrt{16-25x^2}$ is $\frac{x}{2}.{\sqrt{16-25x^2}+\frac{8}{5}.sin^-1(\frac{5x}{4}) +C$

Step-by-step explanation:

We have an equation $\sqrt{16-25x^2}$ and we need to integrate it with respect to x on the interval (-4/5,4/5).

So the integration of a given expression can be done as

$\int\sqrt{16-25x^2}\\=\int\sqrt{(25(16/25)-x^2).dx}\\=\int5\sqrt{(4/5)^2-x^2}\\=5\int\sqrt{(4/5)^2-x^2}$

And we know the property of integral as

$\int(\sqrt{a^2-x^2}=\frac{x}{2}.\sqrt{a^2-x^2}+\frac{a^2}{2}.Sin^{-1}(\frac{x}{a})+C$

The integral becomes

$\int\sqrt{16-25x^2} = 5\int\sqrt{(4/5)-x^2} \\=5(\frac{x}{2}.\sqrt{(4/5)^2-x^2}+\frac{16}{25*.2}.Sin^{-1}{\frac{x}{4/5})}\\=\frac{5x}{2}.\sqrt{(16/25-x^2}+\frac{5.8}{25}Sin^{-1}\frac{5x}{4}$

By solving carely and properly we get

$\int\sqrt{16-25x^2} =$ $\frac{x}{2}.{\sqrt{16-25x^2}+\frac{8}{5}.sin^-1(\frac{5x}{4}) +C$

And it is the required solution to the given problem.