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Valentin Fadeev

Having an inte-great time

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Edited by Valentin Fadeev, Sunday, 16 Jan 2011, 23:18

With the new courses yet to start, hopefully providing fresh material for new posts, I have been spending time going through some excercises from new textbooks.

As integrals have always been my favourite part of calculus, I decided to take down this solution, because it just looks nice. It also illustrates the principle: don't make a substitution, until it becomes obvious.

MathJax failure: TeX parse error: Extra close brace or missing open brace

equation sequence two times x cubed minus three times x squared plus one equals two times x cubed minus two times x squared minus x squared plus one equals two times x squared times open x minus one close minus open x minus one close times open x plus one close equals open x minus one close times open two times x squared minus x minus one close equals open x minus one close times open x squared minus x plus x squared minus one close equals open x minus one close times open x times open x minus one close plus open x minus one close times open x plus one close close equals open x minus one close times open x minus one close times open two times x plus one close equals left parenthesis x minus one times right parenthesis squared times open two times x plus one close

cap i equals Square root of three times integral over negative one divided by two under zero d times x divided by Square root of left parenthesis x minus one times right parenthesis squared times open two times x plus one close

Since negative one divided by two less than or equals x less than or equals zero we have negative three divided by two less than or equals x minus one less than or equals negative one and zero less than or equals two times x plus one less than or equals one , so we need to choose negative sign when taking square root of the quadratic term.

equation sequence cap i equals Square root of three times integral over negative one divided by two under zero d times x divided by Square root of two times x plus one times open one minus x close equals Square root of three times integral over negative one divided by two under zero d of Square root of two times x plus one divided by one minus x

It's now that the substitute equation left hand side two times x plus one equals right hand side t squared becomes an obvious choice.

equation sequence cap i equals Square root of three times integral over zero under one d times t divided by one minus t squared minus one divided by two equals Square root of three times integral over zero under one d times t divided by three divided by two minus t squared divided by two equals two divided by Square root of three times integral over zero under one d times t divided by one minus t squared divided by three equals two times integral over zero under one d of t divided by Square root of three divided by one postfix minus left parenthesis t divided by Square root of three times right parenthesis squared equals two times hyperbolic tangent super negative one of t divided by Square root of three vertical line sub zero super one equals two times hyperbolic tangent super negative one of one divided by Square root of three

 

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Valentin Fadeev

On variable changes in definite integrals

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Edited by Valentin Fadeev, Sunday, 16 Jan 2011, 23:18

Some integrals yield only one type of substitution that really brings them into a convenient form. Any other method would make them more complicated. However, in some cases totally different methods can be applied with equal effect. In case of definite integrals it is of course not necessary to come back to original variable which makes things even easier. Here is one example

integral over zero under one x cubed times d times x divided by Square root of one minus x squared

The most natural way is to apply a trigonometric substitute. We will not consider this method here. Instead an algebraic trick can be employed:

equation sequence integral over zero under one x cubed times d times x divided by Square root of one minus x squared equals integral over zero under one x squared times x times d times x divided by Square root of one minus x squared equals one divided by two times integral over zero under one x squared times d of x squared divided by Square root of one minus x squared equals one divided by two times integral over zero under one t times d times t divided by Square root of one minus t equals

equation sequence negative one divided by two times integral over zero under one open one minus t minus one close times d times t divided by Square root of one minus t equals one divided by two times integral over zero under one d times t divided by Square root of one minus t minus one divided by two times integral over zero under one Square root of one minus t d t equals

negative Square root of one minus t times vertical line sub zero super one plus one divided by three left parenthesis equation sequence one minus t times right parenthesis super three solidus two times vertical line sub zero super one equals one minus one divided by three equals two divided by three

Alternatively we can use integration by parts:

equation sequence integral over zero under one x cubed times d times x divided by Square root of one minus x squared equals integral over zero under one x squared times x times d times x divided by Square root of one minus x squared equals

equation sequence equals integral over zero under one x squared times d of negative Square root of one minus x squared equals negative x squared times Square root of one minus x squared times vertical line sub zero super one plus integral over zero under one Square root of one minus x squared times two times x d x equals

equation left hand side equals right hand side negative two divided by three left parenthesis equation left hand side one minus x squared times right parenthesis super three solidus two times vertical line sub zero super one equals right hand side two divided by three

Or apply an even more exotic treatment:

let x equals one divided by t

equation left hand side integral over zero under one x cubed times d times x divided by Square root of one minus x squared equals right hand side integral over one under normal infinity d times t divided by t super four times Square root of t squared minus one

let t equals hyperbolic cosine of z

equation sequence integral over one under normal infinity d times t divided by t super four times Square root of t squared minus one equals integral over hyperbolic cosine super negative one of one under normal infinity hyperbolic sine of z times d times z divided by left parenthesis hyperbolic cosine of z times right parenthesis super four times hyperbolic sine of z equals integral over hyperbolic cosine super negative one of one under normal infinity d times z divided by left parenthesis hyperbolic cosine of z times right parenthesis super four equals

equation sequence equals integral over hyperbolic cosine super negative one of one under normal infinity left parenthesis left parenthesis hyperbolic cosine of z times right parenthesis squared minus open hyperbolic sine of z times right parenthesis squared close times d times z divided by left parenthesis hyperbolic cosine of z times right parenthesis super four equals

equation sequence equals integral over hyperbolic cosine super negative one of one under normal infinity d times z divided by left parenthesis hyperbolic cosine of z times right parenthesis squared minus integral over hyperbolic cosine super negative one of one under normal infinity left parenthesis hyperbolic tangent of z times right parenthesis squared times d times z divided by left parenthesis hyperbolic cosine of z times right parenthesis squared equals

MathJax failure: TeX parse error: Extra open brace or missing close brace

For

lim over z right arrow normal infinity of hyperbolic tangent of z equals one

and

equation sequence hyperbolic tangent of hyperbolic cosine super negative one of one equals Square root of one minus one divided by left parenthesis hyperbolic cosine of open hyperbolic cosine super negative one of one close times right parenthesis squared equals zero ,

the same result is obtained

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