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- Thread starter nickm4
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HallsofIvy

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[itex]sin^2(t)+ cos^2(t)= 1[/itex] so [itex]\sqrt{1- sin^2(t)}= cos(t)[/itex]

For example, to integrate [itex]\int \sqrt{1- x^2}dx[/itex], let x= sin(t). Then dx= cos(t) dt and [itex]\sqrt{1- x^2}= \sqrt{1- sin^2(t)}= cos(t)[/itex] so the integral becomes [itex]\int cos^2(t) dt[/itex] which can be integrated using the identity [itex]cos^2(t)= (1/2)(1+ cos(2t))[/itex].

For something like [itex]\sqrt{1+ x^2}[/itex] you can divide [itex]sin^2(t)+ cos^2(t)= 1[/itex] by cos(t) to get [itex]tan^2(t)+ 1= sec^2(t)][/itex]

To integrate [itex]\int \sqrt{9+ x^2}dx[/itex], let x= 3tan(t). Then [itex]dx= 3 sec^2(t) dt[/itex] and [itex]\sqrt{9+ x^2}= \sqrt{9+ 9tan^2(t)}= 3\sqrt{1+ tan^2(t)}= 3sec(t)[/itex] so the integral becomes [itex]\int (3 sec(t))(3sec^2(t)dt)= 9\int sec^3(t)dt[/itex]. That can be integrated by writing it as [itex]9\int dt/cos^3(t)= 9\int cos(t)dt/cos^4(t)= 9\int cos(t)dt/(1- sin^2(t))^2[/itex] and using the substitution u= sin(t).

But it is also true that [itex]cosh^2(t)- sinh^2(t)= 1[/itex] or [itex]sinh^2(t)+ 1= cosh^2(t)[/itex] so you could also use the substitution x= 3 sinh(x). Then dx= 3 cosh(x)dx and [itex]\int \sqrt{9+ x^2}dx= 9\int cosh^2(x) dx[/itex]

Just about any calculus text will devote at least a section, if not a chapter, to trig substitutions though hyperbolic substitution are less commonly covered.

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