How do you integrate $\int \frac{\sqrt{x^{2} - a^{2}}}{x^{4}}$ $int sqrt(x^2-a^2)/x^4$ by trigonometric substitution?

Question

How do you integrate $\int \frac{\sqrt{x^{2} - a^{2}}}{x^{4}}$ $int sqrt(x^2-a^2)/x^4$ by trigonometric substitution?

1 Answer

Impact of this question

11657 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-07-26T17:14:24

$\frac{{(x^{2} - a^{2})}^{\frac{3}{2}}}{3 a^{2} x^{3}} + C$ $(x^2-a^2)^(3/2)/(3a^2x^3)+C$ .

Explanation:

We take subst. $x = a sec t, s o, d x = a sec t \cdot tan t d t$ $x=asect, so, dx=asect*tant dt$

Also, $\frac{\sqrt{x^{2} - a^{2}}}{x^{4}}$ $sqrt(x^2-a^2)/x^4$

$= \frac{\sqrt{a^{2} {sec}^{2} t - a^{2}}}{a^{4} {sec}^{4} t}$ $=sqrt(a^2sec^2t-a^2)/(a^4sec^4t)$

$= \frac{a tan t}{a^{4} {sec}^{4} t} = \frac{1}{a^{3}} \cdot \frac{sin t}{cos t} \cdot {cos}^{4} t = \frac{sin t {cos}^{3} t}{a^{3}}$ $=(atant)/(a^4sec^4t)=1/a^3*sint/cost*cos^4t=(sintcos^3t)/a^3$

$:. I=intsqrt(x^2-a^2)/x^4dx=int(sintcos^3t)/a^3*asect*tantdt$

$=1/a^2intsintcos^3t*1/cost*sint/costdt$

$=1/a^2intsin^2tcostdt................(star)$

$=1/a^2int(sint)^2d(sint)$

$=1/a^2*(sint)^3/3$

Now, $sect=x/arArrcost=a/xrArrsint=sqrt(1-a^2/x^2)=sqrt(x^2-a^2)/x$

$:. I=1/(3a^2)*(sqrt(x^2-a^2)/x)^3=(x^2-a^2)^(3/2)/(3a^2x^3)+C$ .

To proceed further from $(star)$ , we can use another subst. $sint=y$ ,

so, $costdt=dy$ , thus giving,

$I=1/a^2inty^2dy=y^3/(3a^2)=(sint)^3/(3a^2)$

Science

Math

Humanities

... and beyond

How do you integrate $\int \frac{\sqrt{x^{2} - a^{2}}}{x^{4}}$ $int sqrt(x^2-a^2)/x^4$ by trigonometric substitution?

1 Answer

Explanation:

Related questions

Impact of this question

Science

Math

Humanities

... and beyond

How do you integrate int sqrt(x^2-a^2)/x^4∫√x2−a2x4int sqrt(x^2-a^2)/x^4 by trigonometric substitution?

1 Answer

Explanation:

Related questions

Impact of this question

How do you integrate $\int \frac{\sqrt{x^{2} - a^{2}}}{x^{4}}$ $int sqrt(x^2-a^2)/x^4$ by trigonometric substitution?