Evaluate the integral. (use c for the constant of integration.) x²/ √49 - x² dx .

Solution:

Given, x²/ √(49 - x²) dx

We have to find the value of the integral using c for the constant of integration.

Now, √(49 - x²) is of the form √(a² - x²) which looks like √(1 - sin²θ)

We know, sin²θ + cos²θ = 1

cos²θ = 1 - sin²θ

So, let a² = 49

a = 7

x = a sinθ

x = 7 sinθ

x² = 49 sin²θ

dx = acosθ dθ

dx = 7 cosθ dθ

Now, √(49 - x²) = √(7)² - (7sinθ)²

= √(49 - 49sin²θ)

= 7√(1 - sin²θ)

= 7√cos²θ

= 7 cosθ

So, \(\int \frac{x^{2}}{\sqrt{49-x^{2}}}=\int \frac{49sin^{2}\theta }{7cos\theta }(7cos\theta )d\theta\)

\(=\int 49sin^{2}\theta \, d\theta\)

\(=49\int sin^{2}\theta \, d\theta\)

We know, \(sin^{2}\theta =\frac{1-cos(2\theta )}{2}\)

So, \(49\int sin^{2}\theta \, d\theta=49\int \frac{1-cos(2\theta )}{2}d\theta \\=\frac{49}{2}\int d\theta -\frac{49}{2}\int cos(2\theta )d\theta \\=\frac{49}{2}\theta -\frac{49}{4}sin(2\theta )+c\)

Consider right triangle, where x/7 = sinθ

θ = arcsin(x/7)

cosθ = \(\frac{\sqrt{49-x^{2}}}{7}\)

sinθ = x/7

Using the identity,

sin(2θ) = 2 sinθ cosθ

So, \(\frac{49}{4}sin(2\theta )=\frac{49}{4}(2sin\theta cos\theta )\\=\frac{49}{4}(2)(\frac{x}{7})(\frac{\sqrt{49-x^{2}}}{7})\\=\frac{x}{2}\sqrt{49-x^{2}}\)

Therefore, \(\int \frac{x^{2}}{\sqrt{49-x^{2}}}=\frac{49}{2}arcsin(\frac{x}{7})+\frac{x}{2}\sqrt{49-x^{2}}+C\).