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I'm not sure that this is the problem, but I think I may not know how to find the $\theta$ value when solving an integral problem with trigonometric substitution.

I got $\frac{\sin^3(\sec^{-1}(x))}{3}+C$ for the answer, but the answer should be, $\frac{1}{3}\frac{(x^2-1)^{3/2}}{x^3}+C$

$$\int\frac{\sqrt{x^2-1}}{x^4}dx$$

Let $x=\sec\theta$

Then $dx=\sec\theta\tan\theta d\theta$

$$\int\frac{\sqrt{\sec^2\theta-1}}{\sec^4\theta}\sec\theta\tan\theta d\theta$$

$$=\int\frac{\sec\theta}{\sec^4\theta}\sqrt{\tan^2\theta}\tan\theta d\theta$$

$$=\int\frac{1}{\sec^3\theta} \tan^2\theta d\theta$$

$$=\int\frac{1}{\sec^3\theta}\frac{\sec^2\theta}{\csc^2\theta}d\theta$$

$$=\int\frac{1}{\sec\theta}\frac{1}{\csc^2\theta}d\theta$$

$$=\int \cos\theta\sin^2\theta d \theta$$

Using $u$-substition, let $u=\sin\theta$

Then $du=\cos\theta d\theta$ and $dx = \frac{1}{\cos\theta}du$

$$\int\cos\theta u^2 \frac{1}{\cos\theta}du$$

$$=\int u^2 du$$

$$=\frac{u^3}{3}+C$$

$$=\frac{\sin^3\theta}{3}+C$$

Since $x=\sec\theta$, $\sec^{-1}(x)=\theta$

$$=\frac{\sin^3(\sec^{-1}(x))}{3}+C$$

What am I doing wrong?

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    $\begingroup$ You did nothing wrong, The two answers are equal. $\endgroup$ – David H Sep 19 '19 at 16:45
  • $\begingroup$ @DavidH Oh! That's weird. How can you determine that? $\endgroup$ – LuminousNutria Sep 19 '19 at 16:46
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    $\begingroup$ $du=\cos\theta\,d\theta,$ not $\cos\theta\,dx$ $\endgroup$ – Thomas Andrews Sep 19 '19 at 16:47
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    $\begingroup$ $\sin(\sec^{-1}(x))=\sin(\cos^{-1}(1/x))=\sqrt{1-1/x^2}=\frac{1}{x}\sqrt{x^2-1}$ $\endgroup$ – Thomas Andrews Sep 19 '19 at 16:48
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    $\begingroup$ Or, use the method of drawing a right triangle with angle $\theta$; since $\sec\theta = x$, we can do this by making the hypotenuse $x$ and the adjacent side $1$. Then the opposite side is $\sqrt{x^2-1}$, so $\sin \theta = \frac{\sqrt{x^2-1}}{x}$. (Though this only really works for $\theta \in (0, \frac{\pi}{2})$, and you would need to work out what the adjustment would need to be for $\theta \in (\frac{\pi}{2}, \pi)$, corresponding to $x < 0$.) $\endgroup$ – Daniel Schepler Sep 19 '19 at 16:57
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You've got the right answer, you've just missed that $$\sin(\sec^{-1}(x))=\sin(\cos^{-1}(1/x))=\sqrt{1-1/x^2}=\frac{1}{x}\sqrt{x^2-1}.$$

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Taking your final answer we can sub back in the original subs $$ \cos (\theta) =\frac{1}{x} $$ We can then use the relationship $$ \sin^2 \theta + \cos^2\theta = 1 = \sin^2 \theta +\frac{1}{x^2} $$ The rest is straight forward.

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Since $x=\sec\theta$, so $\cos\theta=\frac1x$ and hence $\sin\theta=\frac{\sqrt{x^2-1}}{x}$. Thus $$\sin(\sec^{-1}(x))=\frac{(x^2-1)^{3/2}}{x^3}. $$

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  • $\begingroup$ That's the cube of $\sin(\sec^{-1}(x))$ on the right side $\endgroup$ – Thomas Andrews Sep 19 '19 at 16:51

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