I was trying to evaluate the below integral:

$$\int \big[\cos(\csc^{-1}(\tan(\sec^{-1}(\cot(\sin^{-1}(\sec(\cot^{-1}(\csc(\cos^{-1}(x))))))))))\big]^2 dx$$

I managed to simplify it to the below form: $$\int \frac{3x^2-5}{2x^2-3}dx$$I then got to this form: $$\frac{3x}{2}+C-\frac{1}{2}\int\frac{1}{2x^2-3}dx$$ I'm stuck at this point; could anyone show me how to proceed?

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    $\begingroup$ I checked it... you are a monster for composing so many functions!! XD $\endgroup$ Sep 26, 2015 at 5:25
  • $\begingroup$ Take $3$ common in the denominator. And use the traditional trigonometric substitutions. (You can do $\sec\theta=x$) $\endgroup$ Sep 26, 2015 at 5:44
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    $\begingroup$ If I may ask, how did you simplify, in a minute or two, the monster ? $\endgroup$ Sep 26, 2015 at 6:11
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    $\begingroup$ Thank you ! Being now an old man, there is one thing I know for sure : I know very very very little (less than $\epsilon$) ! $\endgroup$ Sep 26, 2015 at 15:39
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    $\begingroup$ Where did you find this exercise :-D $\endgroup$
    – Vajra
    Jun 29, 2021 at 18:53

1 Answer 1


Assuming your simplification is correct (I didn't check), the final step would be handled like this:

$$\int\frac{1}{ax^2 - b}dx = \frac 1b \int\frac{1}{(\sqrt{\frac{a}{b}}x)^2 - 1}dx$$

At this point, you can factor the denominator into $(\sqrt{\frac{a}{b}}x + 1)(\sqrt{\frac{a}{b}}x - 1)$ and use partial fractions.

Alternatively, you could apply the substitution $\sqrt{\frac{a}{b}}x = \cosh y$ which would involve hyperbolic trig functions, if you're comfortable with that.

  • $\begingroup$ Yea thanks a lot! The hyperbolic function is slick; I managed to solve it myself using that before having seen this answer (browsed the net for an appropriate substitution.) Thanks. $\endgroup$ Sep 26, 2015 at 5:44
  • $\begingroup$ @Kugelblitz Glad to help! :) $\endgroup$
    – Deepak
    Sep 26, 2015 at 5:49

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