# Evaluate the following integral. \int \frac{x^2}{\sqrt {361 - x^2}} dx

## Question:

Evaluate the following integral.

{eq}\int \frac{x^2}{\sqrt {361 - x^2}} dx {/eq}

## Integration By Trignometric Substitution:

If the given function contains large terms then those large terms are considered as variables and then the problem is solved.

Let us consider a function f(t) where t is time, then the integral to that function is found out as follows:

{eq}\displaystyle \int f(t) \, dt = F(t) + C {/eq}.

From the above equation, {eq}F(t) {/eq} is the integral and C is an arbitrary constant.

Sum Rule:

If given functions are {eq}f(x) , g(x) {/eq} then

{eq}\displaystyle \int (f(x)+g(x))dx = \int f(x)dx + \int g(x)dx {/eq}

Formulas used are :

{eq}\displaystyle \frac{d\sin x}{dx} = \cos x {/eq}

{eq}\displaystyle \int x^n dx = \frac{x^{n+1}}{n+1} {/eq}

{eq}\displaystyle \sin^2 x = \frac{1-\cos (2x)}{2} {/eq}

Given

{eq}\displaystyle \int \frac{x^2}{\sqrt {361 - x^2}}dx ...................(1) {/eq}

Assume

{eq}\displaystyle x=19\sin u...................(2) {/eq}

On differentiating we get

{eq}\displaystyle dx=19\cos u du...................(3) {/eq}

On substituting equation (2) and equation (3) in equation (1) we get

{eq}\displaystyle =\int 361 \sin^2 u du {/eq}

We can use {eq}\sin^2 x = \frac{1-\cos (3x)}{2} {/eq} in above equation we get

{eq}\displaystyle =\frac{361}{2}\int (1-\cos (2u))du {/eq}

By using sum rule above equation can be written as

{eq}\displaystyle =\frac{361}{2}(\int 1 du - \int \cos (2udu)) {/eq}

{eq}\displaystyle =\frac{361}{2}(u - \frac{\sin (2u)}{2}) {/eq}

Substituting equation (2) in above equation we get

{eq}\displaystyle =\frac{361}{2}(\sin^{-1}(\frac{x}{19}) - \frac{\sin (2\sin^{-1}(\frac{x}{19})}{2}) {/eq}

{eq}\displaystyle \int \frac{x^2}{\sqrt {361 - x^2}} dx=\frac{361}{2}(\sin^{-1}(\frac{x}{19}) - \frac{\sin (2\sin^{-1}(\frac{x}{19})}{2})+c {/eq}

where c is any constant.