How do you integrate $\int \frac{x}{\sqrt{16 - 9 x^{4}}}$ $int x/sqrt(16-9x^4)$ by trigonometric substitution?

Question

How do you integrate $\int \frac{x}{\sqrt{16 - 9 x^{4}}}$ $int x/sqrt(16-9x^4)$ by trigonometric substitution?

1 Answer

Impact of this question

5604 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-03-06T07:33:17

The answer is $= \frac{1}{6} \cdot arcsin (\frac{3}{4} x^{2}) + C$ $=1/6*arcsin(3/4x^2)+ C$

Explanation:

Let's perform some simplification

$16 - 9 x^{4} = 16 (1 - \frac{9}{16} x^{4}) = 16 (1 - {(\frac{3}{4} x^{2})}^{2})$ $16-9x^4=16(1-9/16x^4)=16(1-(3/4x^2)^2)$

Perform the substitution

$sin θ = \frac{3}{4} x^{2}$ $sintheta=3/4x^2$ , $\Rightarrow$ $=>$ , $cos θ d θ = \frac{3}{4} \cdot 2 x d x$ $costheta d theta=3/4*2xdx$

$x d x = \frac{2}{3} cos θ d θ$ $xdx=2/3cos theta d theta$

$16 (1 - {(\frac{3}{4} x^{2})}^{2}) = 16 (1 - {sin}^{2} θ) = 16 {cos}^{2} θ$ $16(1-(3/4x^2)^2)=16(1-sin^2theta)=16cos^2theta$

Therefore,

$\int \frac{x d x}{\sqrt{16 - 9 x^{4}}} = \frac{2}{3} \int \frac{cos θ d θ}{\sqrt{16 {cos}^{2} θ}}$ $int(xdx)/(sqrt(16-9x^4))=2/3int(costheta d theta)/(sqrt(16cos^2theta))$

$= \frac{1}{6} \int d θ$ $=1/6int d theta$

$= \frac{1}{6} θ$ $=1/6theta$

$= \frac{1}{6} \cdot arcsin (\frac{3}{4} x^{2}) + C$ $=1/6*arcsin(3/4x^2)+ C$

Science

Math

Humanities

... and beyond

How do you integrate $\int \frac{x}{\sqrt{16 - 9 x^{4}}}$ $int x/sqrt(16-9x^4)$ by trigonometric substitution?

1 Answer

Explanation:

Related questions

Impact of this question

Science

Math

Humanities

... and beyond

How do you integrate ∫x√16−9x4int x/sqrt(16-9x^4) by trigonometric substitution?

1 Answer

Explanation:

Related questions

Impact of this question

How do you integrate $\int \frac{x}{\sqrt{16 - 9 x^{4}}}$ $int x/sqrt(16-9x^4)$ by trigonometric substitution?