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

Question

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?

Answer 1

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

Answer 2

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.

Answer 3

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

Answer 4

Alternatively, let $J= \int\frac{\sqrt{x^2-1}}{x^2}dx$ and integrate by parts as follows \begin{align}\int\frac{\sqrt{x^2-1}}{x^4}dx =& -\int\frac{(x^2-1)^{3/2}}{x^4}dx+J\\ =&\int (x^2-1)^{3/2}d\left(\frac{1}{3x^3}\right)+J = \frac{({x^2-1})^{3/2}}{3x^3}-J+J \end{align}

Answer 5

Too long for a comment:

One can also evaluate this integral without wrangling with trigonometric substitution by substituting $$u = \frac{\sqrt{x^2 - 1}}{x}, \qquad du = \frac{dx}{x^2 \sqrt{x^2 - 1}} ,$$ which transforms the integral to $$\int u^2 \,du = \frac{1}{3} u^3 + C = \frac{1}{3} \left(\frac{\sqrt{x^2 - 1}}{x}\right)^3 + C = \frac{(x^2 - 1)^\frac32}{3 x^3} + C$$

Answer 6

Letting $y= \frac{1}{x} $ yields

\begin{aligned}\int \frac{\sqrt{x^2-1}}{x^4} d x = & -\int y \sqrt{1-y^2} d y \\ = & \frac{1}{3}\left(1-y^2\right)^{\frac{3}{2}}+C \\ = & \frac{\left(x^2-1\right)^{\frac{3}{2}}}{3 x^3}+C \end{aligned}

Answer 7

No need for any Trigonometric substitution. Your integral is $$\int\frac{\sqrt{x^{2}-1}}{x^{4}}dx$$. Now just take $x^{2}$ common from the numerator. Therefore the integral will become $$\int\frac{x(\sqrt{1-\frac{1}{x^{2}}})}{x^{4}}dx=\int\frac{\sqrt{1-\frac{1}{x^{2}}}}{x^{3}}dx=\frac{-1}{2}\int\frac{-2\sqrt{1-\frac{1}{x^{2}}}}{x^{3}}dx$$. Now substitute $\frac{1}{x^{2}}=u$. Then $\frac{-2}{x^{3}}dx=du$. Therefore finally you will get $$\frac{-1}{2}\int(\sqrt{1-u})du$$. Now I think you can solve very easily.