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Aug
22
accepted Integrating $\frac{1}{(x^4 -1)^2}$
Aug
22
comment Integrating $\frac{1}{(x^4 -1)^2}$
This multiplying and dividing by the derivative of $x^4$ is what my teacher termed as "forced integration". Is there a name for this technique?
Aug
22
comment Integrating $\frac{1}{(x^4 -1)^2}$
@Dr.SonnhardGraubner I'm not sure if I remember correctly, but I was given a hint that it's something about "forced integration". I'll try it using decomposition.
Aug
22
asked Integrating $\frac{1}{(x^4 -1)^2}$
Aug
2
awarded  Great Question
Jun
30
awarded  Nice Question
Jun
27
comment $\cot^{-1}\frac{y}{\sqrt{1-x^2-y^2}} = 2\tan^{-1}\sqrt{\frac{3-4x^2}{4x^2}} - \tan^{-1}\sqrt{\frac{3-4x^2}{x^2}} $
@DavidH Yup, that's correct.
May
19
comment $\cot^{-1}\frac{y}{\sqrt{1-x^2-y^2}} = 2\tan^{-1}\sqrt{\frac{3-4x^2}{4x^2}} - \tan^{-1}\sqrt{\frac{3-4x^2}{x^2}} $
@Mann Converting to $\cos^{-1}$ along with a bit of manipulation does reduce calculation but it's still quite a lot.
May
19
comment $\cot^{-1}\frac{y}{\sqrt{1-x^2-y^2}} = 2\tan^{-1}\sqrt{\frac{3-4x^2}{4x^2}} - \tan^{-1}\sqrt{\frac{3-4x^2}{x^2}} $
@Mann Could you please describe in more detail how the above is useful?
May
19
asked $\cot^{-1}\frac{y}{\sqrt{1-x^2-y^2}} = 2\tan^{-1}\sqrt{\frac{3-4x^2}{4x^2}} - \tan^{-1}\sqrt{\frac{3-4x^2}{x^2}} $
May
19
accepted Sum of the series $\tan^{-1}\frac{4}{4n^2+3}$
May
19
answered Sum of the series $\tan^{-1}\frac{4}{4n^2+3}$
May
19
reviewed Approve Sum of the series $\tan^{-1}\frac{4}{4n^2+3}$
May
19
comment Sum of the series $\tan^{-1}\frac{4}{4n^2+3}$
It was a question on my exam.
May
19
asked Sum of the series $\tan^{-1}\frac{4}{4n^2+3}$
May
15
awarded  Popular Question
Apr
29
awarded  Notable Question
Mar
9
awarded  Popular Question
Dec
25
accepted How to remember trigonometric ratios for allied angles?
Dec
19
awarded  Constituent