inverse trig integration
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BigGlenntheHeavy
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\(\displaystyle \int\frac{dt}{\sqrt{1-4t^{2}}}. \ Let \ u \ = \ 2t, \ \implies \ du \ = \ 2dt\)
\(\displaystyle Hence, \ we \ have: \ \frac{1}{2}\int\frac{du}{\sqrt{1-u^{2}}} \ = \ \frac{arcsin|u|}{2}+C \ = \ \frac{arcsin|2t|}{2}+C\)
\(\displaystyle Hence, \ we \ have: \ \frac{1}{2}\int\frac{du}{\sqrt{1-u^{2}}} \ = \ \frac{arcsin|u|}{2}+C \ = \ \frac{arcsin|2t|}{2}+C\)
BigGlenntheHeavy
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\(\displaystyle OK, \ my \ mistake, \ thought \ t \ was \ a \ + \ sign, \ ergo \ one \ way:\)
\(\displaystyle \int\frac{dt}{t(1-4t^{2})^{1/2}}, \ Let \ u \ = \ (1-4t^{2})^{1/2} \ \implies \ t^{2} \ = \ \frac{1-u^{2}}{4}\)
\(\displaystyle and \ du \ = \ \frac{1}{2}(1-4t^{2})^{-1/2}(-8t)dt \ = \ \frac{-4t}{u}dt \ \implies \ dt \ = \ \frac{u}{-4t}du\)
\(\displaystyle Hence, \ \int\frac{dt}{t(1-4t^{2})^{1/2}} \ = \ \frac{-1}{4}\int\frac{u}{t^{2}u}du \ = \ \frac{-1}{4}\int\frac{du}{t^{2}} \ = \ -\int\frac{du}{1-u^{2}}.\)
\(\displaystyle = \ -\int\bigg[\frac{1}{2(u+1)}-\frac{1}{2(u-1)}\bigg]du \ = \ \frac{1}{2}\int\frac{du}{u-1}-\frac{1}{2}\int\frac{du}{u+1},\)
\(\displaystyle = \ (after \ resub, \ etc.) \ \frac{1}{2}ln\bigg|\frac{(1-4t^{2})^{1/2}-1}{(1-4t^{2})^{1/2}+1}\bigg|, \ Note: \ Absolute \ value \ sign \ is \ necessary.\)
\(\displaystyle Check: \ D_t\bigg[\frac{1}{2}ln\bigg|\frac{(1-4t^{2})^{1/2}-1}{(1-4t^{2})^{1/2}+1}\bigg|\bigg] \ = \ \frac{1}{t(1-4t^{2})^{1/2}}\)
\(\displaystyle \int\frac{dt}{t(1-4t^{2})^{1/2}}, \ Let \ u \ = \ (1-4t^{2})^{1/2} \ \implies \ t^{2} \ = \ \frac{1-u^{2}}{4}\)
\(\displaystyle and \ du \ = \ \frac{1}{2}(1-4t^{2})^{-1/2}(-8t)dt \ = \ \frac{-4t}{u}dt \ \implies \ dt \ = \ \frac{u}{-4t}du\)
\(\displaystyle Hence, \ \int\frac{dt}{t(1-4t^{2})^{1/2}} \ = \ \frac{-1}{4}\int\frac{u}{t^{2}u}du \ = \ \frac{-1}{4}\int\frac{du}{t^{2}} \ = \ -\int\frac{du}{1-u^{2}}.\)
\(\displaystyle = \ -\int\bigg[\frac{1}{2(u+1)}-\frac{1}{2(u-1)}\bigg]du \ = \ \frac{1}{2}\int\frac{du}{u-1}-\frac{1}{2}\int\frac{du}{u+1},\)
\(\displaystyle = \ (after \ resub, \ etc.) \ \frac{1}{2}ln\bigg|\frac{(1-4t^{2})^{1/2}-1}{(1-4t^{2})^{1/2}+1}\bigg|, \ Note: \ Absolute \ value \ sign \ is \ necessary.\)
\(\displaystyle Check: \ D_t\bigg[\frac{1}{2}ln\bigg|\frac{(1-4t^{2})^{1/2}-1}{(1-4t^{2})^{1/2}+1}\bigg|\bigg] \ = \ \frac{1}{t(1-4t^{2})^{1/2}}\)
BigGlenntheHeavy
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\(\displaystyle Another \ way \ (probably \ the \ easiest):\)
\(\displaystyle \int\frac{dt}{t(1-4t^{2})^{1/2}}, \ Let \ 2t \ = \ sin(\theta) \ \implies \ t \ = \ \frac{sin(\theta)}{2} \ and \ dt \ = \ \frac{cos(\theta)}{2}d\theta.\)
\(\displaystyle Hence, \ we \ have \ \int\frac{cos(\theta)}{sin(\theta)cos(\theta)}d\theta \ = \ \int csc(\theta)d\theta \ = \ -ln|csc(\theta)+cot(\theta)|+C\)
\(\displaystyle Hence, \ we \ have \ \int\frac{dt}{t(1-4t^{2})^{1/2}} \ = \ -ln\bigg|\frac{1+(1-4t^{2})^{1/2}}{2t}\bigg|+C, \ -1/2 \ \le \ \ t \ \le \ 1/2, \ t \ \ne \ 0.\)
\(\displaystyle Check: \ D_t\bigg[-ln\bigg|\frac{1+(1-4t^{2})^{1/2}}{2t}\bigg|+C\bigg] \ = \ \frac{1}{t(1-4t^{2})^{1/2}}.\)
\(\displaystyle \int\frac{dt}{t(1-4t^{2})^{1/2}}, \ Let \ 2t \ = \ sin(\theta) \ \implies \ t \ = \ \frac{sin(\theta)}{2} \ and \ dt \ = \ \frac{cos(\theta)}{2}d\theta.\)
\(\displaystyle Hence, \ we \ have \ \int\frac{cos(\theta)}{sin(\theta)cos(\theta)}d\theta \ = \ \int csc(\theta)d\theta \ = \ -ln|csc(\theta)+cot(\theta)|+C\)
\(\displaystyle Hence, \ we \ have \ \int\frac{dt}{t(1-4t^{2})^{1/2}} \ = \ -ln\bigg|\frac{1+(1-4t^{2})^{1/2}}{2t}\bigg|+C, \ -1/2 \ \le \ \ t \ \le \ 1/2, \ t \ \ne \ 0.\)
\(\displaystyle Check: \ D_t\bigg[-ln\bigg|\frac{1+(1-4t^{2})^{1/2}}{2t}\bigg|+C\bigg] \ = \ \frac{1}{t(1-4t^{2})^{1/2}}.\)
BigGlenntheHeavy
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\(\displaystyle Right \ Aladdin, \ look \ at \ the \ integral. \ If \ we \ let f(x) \ = \ \frac{1}{t(1-4t^{2})^{1/2}},\)
\(\displaystyle then \ (1-4t^{2}) \ >0, \ Why? \ This \ implies \ 4t^{2} \ < \ 1, \ implies \ t^{2} \ < \ \frac{1}{4},\)
\(\displaystyle implies \ |t| \ < \ \frac{1}{2}, \ hence \ -\frac{1}{2} \ < \ t \ < \ \frac{1}{2}, \ t \ \ne \ 0\)
\(\displaystyle So, \ the \ solution \ is \ correct \ providing \ it \ is \ within \ this \ domain.\)
\(\displaystyle Just \ remember \ Aladdin, \ division \ by \ zero \ is \ a \ no-no.\)
\(\displaystyle then \ (1-4t^{2}) \ >0, \ Why? \ This \ implies \ 4t^{2} \ < \ 1, \ implies \ t^{2} \ < \ \frac{1}{4},\)
\(\displaystyle implies \ |t| \ < \ \frac{1}{2}, \ hence \ -\frac{1}{2} \ < \ t \ < \ \frac{1}{2}, \ t \ \ne \ 0\)
\(\displaystyle So, \ the \ solution \ is \ correct \ providing \ it \ is \ within \ this \ domain.\)
\(\displaystyle Just \ remember \ Aladdin, \ division \ by \ zero \ is \ a \ no-no.\)
BigGlenntheHeavy
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\(\displaystyle Look, \ \sqrt{1-4t^2}, \ then \ 1-4t^2 \ \ge \ 0 \ usually.\)
\(\displaystyle However \ if \ \sqrt{1-4t^{2}} \ is \ the \ denominator \ of \ a \ fraction, \ then \ {1-4t^{2}} \ > \ 0\)
\(\displaystyle as \ if \ it \ equal \ zero, \ then \ we \ would \ have \ division \ by \ zero, \ a \ no-no.\)
\(\displaystyle However \ if \ \sqrt{1-4t^{2}} \ is \ the \ denominator \ of \ a \ fraction, \ then \ {1-4t^{2}} \ > \ 0\)
\(\displaystyle as \ if \ it \ equal \ zero, \ then \ we \ would \ have \ division \ by \ zero, \ a \ no-no.\)
BigGlenntheHeavy
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\(\displaystyle Another \ thing \ Aladdin, \ we \ were \ given \ to \ find \ the \ integral \ of \ a \ function, \ which \ we \ did.\)
\(\displaystyle The \ problem \ said \ nothing \ about \ the \ domain \ in \ which \ this \ integral \ will \ work.\)
\(\displaystyle You \ professor \ is \ wrong: \ as \ we \ were \ ask \ to \ find \ the \ antiderivative \ of \ f(x), \ not \ its \ domain\)
\(\displaystyle \ or \ range.\)
\(\displaystyle The \ problem \ said \ nothing \ about \ the \ domain \ in \ which \ this \ integral \ will \ work.\)
\(\displaystyle You \ professor \ is \ wrong: \ as \ we \ were \ ask \ to \ find \ the \ antiderivative \ of \ f(x), \ not \ its \ domain\)
\(\displaystyle \ or \ range.\)
BigGlenntheHeavy
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\(\displaystyle Here \ is \ the \ proof; \ send \ it \ to \ your \ professor.\)
\(\displaystyle \int\frac{dt}{t(1-4t^{2})^{1/2}} \ = \ -ln\bigg|\frac{1+(1-4t^{2})^{1/2}}{2t}\bigg|+C\)
\(\displaystyle Check: \ D_t\bigg[-ln\bigg|\frac{1+(1-4t^{2})^{1/2}}{2t}\bigg|+C\bigg] \ = \ \frac{1}{t(1-4t^{2})^{1/2}}.\)
\(\displaystyle \int\frac{dt}{t(1-4t^{2})^{1/2}} \ = \ -ln\bigg|\frac{1+(1-4t^{2})^{1/2}}{2t}\bigg|+C\)
\(\displaystyle Check: \ D_t\bigg[-ln\bigg|\frac{1+(1-4t^{2})^{1/2}}{2t}\bigg|+C\bigg] \ = \ \frac{1}{t(1-4t^{2})^{1/2}}.\)
As Glenn pointed out, this is a function with domain (-1/2,0)U(0,1/2). That is as general as it gets for this f in the Reals. Glenn's solution is the anti-derivative for f on any subset of this domain. The Definite Integral will converge for any bounds, say a to b, in the domain of the function as long as a and b have the same sign. The non-removable discontinuity at 0 is what might be causing the fuss with your professor.