What did I get wrong when solving ∫x2−1x4dx? Let x=sec⁡θ Then dx=sec⁡θtan⁡θdθ ∫sec2θ−1sec4θsec⁡θtan⁡θdθ =∫sec⁡θsec4θtan2θtan⁡θdθ =∫1sec3θtan2θdθ =∫1sec3θsec2θcsc2θdθ =∫1sec⁡θ1csc2θdθ =∫cos⁡θsin2θdθ Using u-substition, let u=sin⁡θ Then du=cos⁡θdθ and dx=1cos⁡θdu ∫cos⁡θu21cos⁡θdu =∫u2du =u33+C =sin3θ3+C Since x=sec⁡θ,sec−1(x)=θ =sin3(sec−1(x))3+C

Question

What did I get wrong when solving ∫x2−1x4dx?  Let x=sec⁡θ   Then dx=sec⁡θtan⁡θdθ  ∫sec2θ−1sec4θsec⁡θtan⁡θdθ   =∫sec⁡θsec4θtan2θtan⁡θdθ   =∫1sec3θtan2θdθ   =∫1sec3θsec2θcsc2θdθ   =∫1sec⁡θ1csc2θdθ   =∫cos⁡θsin2θdθ   Using u-substition, let u=sin⁡θ   Then du=cos⁡θdθ  and  dx=1cos⁡θdu  ∫cos⁡θu21cos⁡θdu   =∫u2du   =u33+C   =sin3θ3+C   Since x=sec⁡θ,sec−1(x)=θ  =sin3(sec−1(x))3+C

Aaron Hughes · Accepted Answer

You made a mistake because you already knew the right answer.sin⁡(sec−1⁡(x))=sin⁡(cos−1⁡(1/x))=1−1/x2=1xx2−1.&amp;nbsp;

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

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