# What did I get wrong when solving $\int\frac{\sqrt{x^2-1}}{x^4}dx$?

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?

• You did nothing wrong, The two answers are equal. – David H Sep 19 '19 at 16:45
• @DavidH Oh! That's weird. How can you determine that? – LuminousNutria Sep 19 '19 at 16:46
• $du=\cos\theta\,d\theta,$ not $\cos\theta\,dx$ – Thomas Andrews Sep 19 '19 at 16:47
• $\sin(\sec^{-1}(x))=\sin(\cos^{-1}(1/x))=\sqrt{1-1/x^2}=\frac{1}{x}\sqrt{x^2-1}$ – Thomas Andrews Sep 19 '19 at 16:48
• 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 $x < 0$.) – Daniel Schepler Sep 19 '19 at 16:57

You've got the right answer, you've just missed that $$\sin(\sec^{-1}(x))=\sin(\cos^{-1}(1/x))=\sqrt{1-1/x^2}=\frac{1}{x}\sqrt{x^2-1}.$$
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.
Since $$x=\sec\theta$$, so $$\cos\theta=\frac1x$$ and hence $$\sin\theta=\frac{\sqrt{x^2-1}}{x}$$. Thus $$\sin(\sec^{-1}(x))=\frac{(x^2-1)^{3/2}}{x^3}.$$
• That's the cube of $\sin(\sec^{-1}(x))$ on the right side – Thomas Andrews Sep 19 '19 at 16:51