Question

Integrate x²dx/(9-x²)½ using trigonometric substitution

Answer 1

$I=\int \frac{x^2}{\left(9-x^2\right)^{\frac{1}{2}}}dx$

Let $x = 3 sin(z)$

$\Rightarrow dx=3cos(z)dz$

substituting these value in $I$

$I=\int \frac{9sin^2\left(z\right)}{\left(9-9sin^2(z)\right)^{\frac{1}{2}}}\cdot 3cos\left(z\right)dz$

$=\int \frac{9sin^2\left(z\right)}{\left(9\left(1-sin^2\left(z\right)\right)\right)^{\frac{1}{2}}}\cdot 3cos\left(z\right)dz$

$=\int\frac{9sin^2\left(z\right)}{\left(9cos^2\left(z\right)\right)^{\frac{1}{2}}}\cdot 3cos\left(z\right)dz$

$=\int\frac{9sin^2\left(z\right)}{3cosz}\cdot 3cos\left(z\right)dz$

$=9\int sin^2(z)dz$

$=9 \int \:\frac{1-cos(2z)}{2}dz$ as $cos(2z)=1-2sin^2(z)$

$=\frac{9}{2}\int \:\left(1-cos\left(2z\right)\right)dz$

$=\frac{9}{2}\left(z-\frac{sin\left(2z\right)}{2}\right)+C$

We assumed $x=3sin(z)$

$\Rightarrow$ $sin(z)=\frac{x}3$

$z=sin^{-1}(\frac{x}{3})$

substituting back these value

$I=\frac{9}{2}\left(sin^{-1}\left(\frac{x}{3}\right)-\frac{1}{2}sin\left(2sin^{-1}\left(\frac{x}{3}\right)\right)\right)+C$

where C is the constant of integration.