9x-2-49-x-3-dx-

Question Number 212456 by Hanuda354 last updated on 14/Oct/24

\int \frac{\sqrt{9 x^{2} - 49}}{x^{3}} d x = ?

Answered by Sutrisno last updated on 14/Oct/24

\int \frac{\sqrt{{(3 x)}^{2} - 7^{2}}}{x^{3}} d x

3 x = 7 s e c θ \to d x = \frac{7}{3} s e c θ t a n θ d θ

= \int \frac{\sqrt{{(7 s e c θ)}^{2} - 7^{2}}}{{(\frac{7}{3} s e c θ)}^{3}} . \frac{7}{3} s e c θ t a n θ d θ

= \int \frac{7 t a n θ}{\frac{343}{27} {s e c}^{3} θ} . \frac{7}{3} s e c θ t a n θ d θ

= \frac{9}{7} \int \frac{{t a n}^{2} θ}{{s e c}^{2} θ} d θ

= \frac{9}{7} \int {s i n}^{2} θ d θ

= \frac{9}{7} \int \frac{1 - c o s 2 θ}{2} d θ

= \frac{9}{14} (θ - \frac{1}{2} s i n 2 θ)

= \frac{9}{14} (θ - s i n θ c o s θ) + c

= \frac{9}{14} ({s e c}^{- 1} (\frac{3 x}{7}) - \frac{\sqrt{9 x^{2} - 49}}{3 x} . \frac{7}{3 x}) + c

= \frac{9}{14} ({s e c}^{- 1} (\frac{3 x}{7}) - \frac{7 \sqrt{9 x^{2} - 49}}{9 x^{2}}) + c

Commented by Hanuda354 last updated on 14/Oct/24

Thanks

Answered by Ghisom last updated on 14/Oct/24

\int \frac{\sqrt{9 x^{2} - 49}}{x^{3}} d x =

[t = \arcsin \frac{14 \sqrt{9 x^{2} - 49}}{x^{2}} \to d x = - \frac{x \sqrt{9 x^{2} - 49}}{14} d t]

= - \frac{9}{28} \int (1 + \cos t) d t = - \frac{9}{28} (t + \sin t) =

= - \frac{\sqrt{9 x^{2} - 49}}{2 x^{2}} - \frac{9}{28} \arcsin \frac{14 \sqrt{9 x^{2} - 49}}{x^{2}} + C

Commented by Hanuda354 last updated on 14/Oct/24

Thank you