How do you integrate $\int \frac{1}{x^{2} \sqrt{16 x^{2} - 9}}$ $int 1/(x^2sqrt(16x^2-9))$ by trigonometric substitution?

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How do you integrate $\int \frac{1}{x^{2} \sqrt{16 x^{2} - 9}}$ $int 1/(x^2sqrt(16x^2-9))$ by trigonometric substitution?

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Answer 1 · 2018-03-03T18:22:45

$\frac{\sqrt{16 x^{2} - 9}}{9 x} + C$ $sqrt(16x^2-9)/(9x)+C$ .

Explanation:

Suppose that, $I = \int \frac{1}{x^{2} \sqrt{16 x^{2} - 9}} d x = \int \frac{1}{x^{2} \sqrt{{(4 x)}^{2} - 3^{2}}} d x$ $I=int1/(x^2sqrt(16x^2-9))dx=int1/(x^2sqrt{(4x)^2-3^2})dx$ .

We subst. $4 x = 3 sec y, so that, 4 d x = 3 sec y tan y d y$ $4x=3secy," so that, "4dx=3secytanydy$ .

$:. I=int1/{(3/4*secy)^2sqrt(9sec^2y-9)}*(3/4*secytany)dy$ ,

$=16/9*1/4int1/(sec^2ytany)*secytanydy$ ,

$=4/9int1/secydy$ ,

$=4/9intcosydy$ ,

$=4/9siny$ ,

$=4/9sqrt(1-cos^2y)$ ,

$=4/9sqrt(1-1/sec^2y)$ ,

$=4/9sqrt{1-1/(4/3*x)^2}$ ,

$=4/9sqrt{1-9/(16x^2)}$ .

$rArr I=sqrt(16x^2-9)/(9x)+C$ .

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How do you integrate $\int \frac{1}{x^{2} \sqrt{16 x^{2} - 9}}$ $int 1/(x^2sqrt(16x^2-9))$ by trigonometric substitution?

1 Answer

Explanation:

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How do you integrate int 1/(x^2sqrt(16x^2-9))∫1x2√16x2−9int 1/(x^2sqrt(16x^2-9)) by trigonometric substitution?

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How do you integrate $\int \frac{1}{x^{2} \sqrt{16 x^{2} - 9}}$ $int 1/(x^2sqrt(16x^2-9))$ by trigonometric substitution?