The Notes: `int x^2 /(3 + 4x - 4x^2)^(3/2) dx` Evaluate the integral

Wednesday, June 8, 2011

$\displaystyle\int\frac{{{x}}^{{2}}}{{{\left({3}+{4}{x}-{4}{{x}}^{{2}}\right)}}^{{\frac{{3}}{{2}}}}}{\left.{d}{x}\right.}$ Evaluate the integral

$\displaystyle\int\frac{{{x}}^{{2}}}{{{\left({3}+{4}{x}-{4}{{x}}^{{2}}\right)}}^{{\frac{{3}}{{2}}}}}{\left.{d}{x}\right.}$

Let's rewrite the denominator of the integrand,

$\displaystyle=\int\frac{{{x}}^{{2}}}{{{\left(-{\left({4}{{x}}^{{2}}-{4}{x}+{1}\right)}+{4}\right)}}^{{\frac{{3}}{{2}}}}}{\left.{d}{x}\right.}$

$\displaystyle=\int\frac{{{x}}^{{2}}}{{{\left({\left({{2}}^{{2}}-{{\left({2}{x}-{1}\right)}}^{{2}}\right)}\right)}}^{{\frac{{3}}{{2}}}}}{\left.{d}{x}\right.}$

Now let's use the integral substitution,

Let $\displaystyle{2}{x}-{1}={2}{\sin{{\left(\theta\right)}}}$

$\displaystyle\Rightarrow{2}{x}={1}+{2}{\sin{{\left(\theta\right)}}}$

$\displaystyle\Rightarrow{x}=\frac{{{1}+{2}{\sin{{\left(\theta\right)}}}}}{{2}}$

$\displaystyle{\left.{d}{x}\right.}=\frac{{1}}{{2}}{\left({2}{\cos{{\left(\theta\right)}}}\right)}{d}\theta$

$\displaystyle{\left.{d}{x}\right.}={\cos{{\left(\theta\right)}}}{d}\theta$

Plug the above in the integral,

$\displaystyle=\int\frac{{{\left(\frac{{{1}+{2}{\sin{{\left(\theta\right)}}}}}{{2}}\right)}}^{{2}}}{{{\left({{2}}^{{2}}-{{2}}^{{2}}{{\sin}}^{{2}}{\left(\theta\right)}\right)}}^{{\frac{{3}}{{2}}}}}{\cos{{\left(\theta\right)}}}{d}\theta$

$\displaystyle=\int\frac{{1}}{{4}}\frac{{{{\left({1}+{2}{\sin{{\left(\theta\right)}}}\right)}}^{{2}}{\cos{{\left(\theta\right)}}}}}{{{\left({{2}}^{{2}}{\left({1}-{{\sin}}^{{2}}{\left(\theta\right)}\right)}\right)}}^{{\frac{{3}}{{2}}}}}{d}\theta$

$\displaystyle=\frac{{1}}{{4}}\int\frac{{{{\left({1}+{2}{\sin{{\left(\theta\right)}}}\right)}}^{{2}}{\cos{{\left(\theta\right)}}}}}{{{{\left({{2}}^{{2}}\right)}}^{{\frac{{3}}{{2}}}}{{\left({1}-{{\sin}}^{{2}}{\left(\theta\right)}\right)}}^{{\frac{{3}}{{2}}}}}}{d}\theta$

$\displaystyle=\frac{{1}}{{4}}\int\frac{{{{\left({1}+{2}{\sin{{\left(\theta\right)}}}\right)}}^{{2}}{\cos{{\left(\theta\right)}}}}}{{{{2}}^{{3}}{{\left({1}-{{\sin}}^{{2}}{\left(\theta\right)}\right)}}^{{\frac{{3}}{{2}}}}}}{d}\theta$

Now use the identity: $\displaystyle{1}-{{\sin}}^{{2}}{\left({x}\right)}={{\cos}}^{{2}}{\left({x}\right)}$

$\displaystyle=\frac{{1}}{{32}}\int\frac{{{{\left({1}+{2}{\sin{{\left(\theta\right)}}}\right)}}^{{2}}{\cos{{\left(\theta\right)}}}}}{{{\left({{\cos}}^{{2}}{\left(\theta\right)}\right)}}^{{\frac{{3}}{{2}}}}}{d}\theta$

$\displaystyle=\frac{{1}}{{32}}\int\frac{{{\left({1}+{4}{\sin{{\left(\theta\right)}}}+{4}{{\sin}}^{{2}}{\left(\theta\right)}\right)}{\cos{{\left(\theta\right)}}}}}{{{{\cos}}^{{3}}{\left(\theta\right)}}}{d}\theta$

$\displaystyle=\frac{{1}}{{32}}\int\frac{{{1}+{4}{\sin{{\left(\theta\right)}}}+{4}{{\sin}}^{{2}}{\left(\theta\right)}}}{{{{\cos}}^{{2}}{\left(\theta\right)}}}{d}\theta$

$\displaystyle=\frac{{1}}{{32}}\int{\left(\frac{{1}}{{{{\cos}}^{{2}}{\left(\theta\right)}}}+\frac{{{4}{\sin{{\left(\theta\right)}}}}}{{{{\cos}}^{{2}}{\left(\theta\right)}}}+\frac{{{4}{{\sin}}^{{2}}{\left(\theta\right)}}}{{{{\cos}}^{{2}}{\left(\theta\right)}}}{d}\theta\right.}$

$\displaystyle=\frac{{1}}{{32}}\int{\left({{\sec}}^{{2}}{\left(\theta\right)}+{4}{\tan{{\left(\theta\right)}}}{\sec{{\left(\theta\right)}}}+{4}{{\tan}}^{{2}}{\left(\theta\right)}\right)}{d}\theta$

Now use the identity: $\displaystyle{{\tan}}^{{2}}{\left({x}\right)}={{\sec}}^{{2}}{\left({x}\right)}-{1}$

$\displaystyle=\frac{{1}}{{32}}\int{\left({{\sec}}^{{2}}{\left(\theta\right)}+{4}{\tan{{\left(\theta\right)}}}{\sec{{\left(\theta\right)}}}+{4}{\left({{\sec}}^{{2}}{\left(\theta\right)}-{1}\right)}{d}\theta\right.}$

$\displaystyle=\frac{{1}}{{32}}\int{\left({5}{{\sec}}^{{2}}{\left(\theta\right)}+{4}{\tan{{\left(\theta\right)}}}{\sec{{\left(\theta\right)}}}-{4}\right)}{d}\theta$

Now use the standard integrals,

$\displaystyle\int{{\sec}}^{{2}}{\left({x}\right)}{\left.{d}{x}\right.}={\tan{{\left({x}\right)}}}+{C}$

$\displaystyle\int{\sec{{\left({x}\right)}}}{\tan{{\left({x}\right)}}}{\left.{d}{x}\right.}={\sec{{\left({x}\right)}}}+{C}$

$\displaystyle=\frac{{1}}{{32}}{\left({5}{\tan{{\left(\theta\right)}}}+{4}{\sec{{\left(\theta\right)}}}-{4}\theta\right)}+{C}$

We have used the integral substitution $\displaystyle{2}{x}-{1}={2}{\sin{{\left(\theta\right)}}}$

$\displaystyle\Rightarrow{\sin{{\left(\theta\right)}}}=\frac{{{2}{x}-{1}}}{{2}}$

$\displaystyle\theta={\arcsin{{\left(\frac{{{2}{x}-{1}}}{{2}}\right)}}}$

Now let's find the $\displaystyle{\tan{{\left(\theta\right)}}}$ and $\displaystyle{\sec{{\left(\theta\right)}}}$ using the right triangle with angle $\displaystyle\theta$ and opposite side (2x-1) and hypotenuse as 2,

Use pythagorean identity to find the adjacent side A:

$\displaystyle{{A}}^{{2}}+{{\left({2}{x}-{1}\right)}}^{{2}}={{2}}^{{2}}$

$\displaystyle{{A}}^{{2}}+{4}{{x}}^{{2}}-{4}{x}+{1}={4}$

$\displaystyle{{A}}^{{2}}={4}-{1}+{4}{x}-{4}{{x}}^{{2}}={3}+{4}{x}-{4}{{x}}^{{2}}$

$\displaystyle{A}=\sqrt{{{3}+{4}{x}-{4}{{x}}^{{2}}}}$

$\displaystyle{\tan{{\left(\theta\right)}}}=\frac{{{2}{x}-{1}}}{{\sqrt{{{3}+{4}{x}-{4}{{x}}^{{2}}}}}}$

$\displaystyle{\sec{{\left(\theta\right)}}}=\frac{{2}}{{\sqrt{{{3}+{4}{x}-{4}{{x}}^{{2}}}}}}$

Now plug these in the above solution,

$\displaystyle=\frac{{1}}{{32}}{\left({5}\cdot\frac{{{2}{x}-{1}}}{{\sqrt{{{3}+{4}{x}-{4}{{x}}^{{2}}}}}}+{4}\cdot\frac{{2}}{{\sqrt{{{3}+{4}{x}-{4}{{x}}^{{2}}}}}}-{4}{\arcsin{{\left(\frac{{{2}{x}-{1}}}{{2}}\right)}}}\right)}+{C}$

$\displaystyle=\frac{{1}}{{32}}{\left(\frac{{{10}{x}-{5}+{8}}}{\sqrt{{{3}+{4}{x}-{4}{{x}}^{{2}}}}}-{4}{\arcsin{{\left(\frac{{{2}{x}-{1}}}{{2}}\right)}}}\right)}+{C}$

$\displaystyle=\frac{{1}}{{32}}{\left(\frac{{{10}{x}+{3}}}{\sqrt{{{3}+{4}{x}-{4}{{x}}^{{2}}}}}-{4}{\arcsin{{\left(\frac{{{2}{x}-{1}}}{{2}}\right)}}}\right)}+{C}$

The Notes

Wednesday, June 8, 2011

$\displaystyle\int\frac{{{x}}^{{2}}}{{{\left({3}+{4}{x}-{4}{{x}}^{{2}}\right)}}^{{\frac{{3}}{{2}}}}}{\left.{d}{x}\right.}$ Evaluate the integral

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