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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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