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I'm having trouble with this trig substitution problem. Please help me out.

$$\int{\frac{ \sqrt{9x^2-289}}{x}}\;dx$$

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Hi Sarah welcome to mathematics stackexchange. We made some edits to your post but weren't entirely clear as to whether the square root applied to the denominator or not. Please can you check and correct if we got it wrong? Also, it would help if you could supply some more context as to the approach you've tried, e.g. what trig substitutions have you tried? – TooTone Feb 14 '14 at 23:45
Great thanks for the edit :), sorry i'm not sure how to approach this problem, I tried substituting the sqrt area with secant by trig sub but i'm not sure how to go from there.. – user128885 Feb 14 '14 at 23:55

Hint: $$\int{\frac{ \sqrt{9x^2-289}}{x}}\;dx = \int\frac{\sqrt{(3x)^2 - (17)^2}}{x}$$

Now put $$3x = 17\sec\theta \implies x = \frac{17}3 \sec \theta \implies dx = \frac {17}{3} \tan \theta \sec \theta\, d\theta,\; \\ \theta = \sec^{-1}\left(\frac {3x}{17}\right)\;\text{ and }\;\sqrt{9x^2 - 289} = 17\sqrt{\sec^2\theta - 1} = 17\sqrt{\tan^2\theta} = 17\tan \theta$$

If you proceed with the substitution correctly, you'll arrive at $$17 \int \tan^2 \theta \,d\theta = 17\int(\sec^2\theta - 1)\,d\theta = 17\tan \theta - 17\theta + C$$

Now, we determined above both $17 \tan\theta$ and $\theta$, each in terms of $x$, based on your original substitution. Using these "back-substitutions", you are good to go!

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This method looks a bit different because I prefer to perform inverse trig substitutions directly (instead of indirectly, as is more conventional). It's more work but I allows me to see what inverse trig substitution "really" is, and takes the magic out that $dx=f(\theta)d\theta$ thingy.

First off, transform the square root into the form $\sqrt{u^2-1}$ $$\begin{array}{lll} \int \frac{\sqrt{9x^2-289}}{x}dx &=& 17\int \frac{\sqrt{(\frac{3x}{17})^2-1}}{\frac{17}{3}\cdot\frac{3x}{17}}dx\\ &=& 3\int \frac{\sqrt{(\frac{3x}{17})^2-1}}{\frac{3x}{17}}dx\\ &=& 3\int \frac{\sqrt{(\frac{3x}{17})^2-1}}{\frac{3x}{17}}dx\cdot\frac{\frac{d(\frac{3x}{17})}{dx}}{\frac{d(\frac{3x}{17})}{dx}}\text{ notice how I multiply by 1}\\ &=& 3\int \frac{\sqrt{(\frac{3x}{17})^2-1}}{\frac{3x}{17}}\frac{d(\frac{3x}{17})}{\frac{3}{17}}\\ &=& 17\int \frac{\sqrt{(\frac{3x}{17})^2-1}}{\frac{3x}{17}}d(\frac{3x}{17})\\ &=& 17\int \frac{\sqrt{u^2-1}}{u}du, \text{ where }u=\frac{3x}{17}\\ \end{array}$$ Next find the appropriate inverse trigonometric substitution $$\theta=\sec^{-1}u$$ $$\sec \theta = u$$ $$\frac{d\sec \theta}{d\theta} = \frac{du}{d\theta}$$ $$\sec\theta\tan\theta = \frac{du}{d\theta}$$ $$\frac{d\theta}{du} = \frac{1}{\sec\theta\tan\theta}$$ $$\frac{d\theta}{du} = \frac{1}{\sec\theta\sqrt{\sec^2\theta-1}} = \frac{1}{\sec(\sec^{-1}u)\sqrt{\sec^2(\sec^{-1}u)-1}}=\frac{1}{u\sqrt{u^2-1}}$$

Now apply the substitution to the integral

$$\begin{array}{lll} 17\int \frac{\sqrt{u^2-1}}{u}du &=& 17\int \frac{\sqrt{u^2-1}}{u}du\cdot\frac{\frac{d\theta}{du}}{\frac{d\theta}{du}}\\ &=& 17\int \frac{\sqrt{u^2-1}}{u}\cdot\frac{1}{\frac{1}{u\sqrt{u^2-1}}}d\theta\\ &=& 17\int \frac{\sqrt{u^2-1}}{u}\cdot\frac{1}{\frac{1}{u\sqrt{u^2-1}}}d\theta\\ &=& 17\int (u^2-1) d\theta\\ &=& 17\int (\sec^2(\sec^{-1}u)-1) d\theta\\ &=& 17\int (\sec^2\theta-1) d\theta\\ &=& \dots\\ \end{array}$$

Hopefully, you try a few this way before going back to that $dx=f(\theta)d\theta$ thingy. :)

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