2
$\begingroup$

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?

$\endgroup$
7
  • 1
    $\begingroup$ 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
  • 1
    $\begingroup$ $du=\cos\theta\,d\theta,$ not $\cos\theta\,dx$ $\endgroup$ – Thomas Andrews Sep 19 '19 at 16:47
  • 1
    $\begingroup$ $\sin(\sec^{-1}(x))=\sin(\cos^{-1}(1/x))=\sqrt{1-1/x^2}=\frac{1}{x}\sqrt{x^2-1}$ $\endgroup$ – Thomas Andrews Sep 19 '19 at 16:48
  • 1
    $\begingroup$ 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
2
$\begingroup$

You've got the right answer, you've just missed that $$\sin(\sec^{-1}(x))=\sin(\cos^{-1}(1/x))=\sqrt{1-1/x^2}=\frac{1}{x}\sqrt{x^2-1}.$$

$\endgroup$
1
$\begingroup$

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.

$\endgroup$
1
$\begingroup$

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

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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.