# The addition-subtraction puzzle (calculus)

• Jan 12th 2011, 12:02 PM
wonderboy1953
The addition-subtraction puzzle (calculus)
In my easy calculus puzzle, I've submitted $\dfrac{1}{\sin(x)cos(x)}$ (in another form) and I pointed out how by letting 1 = $sin^2x + cos^2x$ would lead to a solution. The formula can be rewritten as 1 = $(1 -cox^2x) + cos^2x$ for the numerator of the first fraction which of course leads to the same answer.

That method I refer to as the addition-subtraction method or ASM for short (which is applied to the numerators of fractions). This is a method that should be taught in calculus classes and demonstrated in texts, but unfortunately isn't. It's a powerful method for simplifying fractions and I'll now give you another example in the form of an easy puzzle.

Using ASM, transform $\dfrac{1}{x(1 + x^2)}$ into two fractions and then go on to integrate those two fractions.
• Jan 14th 2011, 05:59 AM
Pranas
Quote:

Originally Posted by wonderboy1953
In my easy calculus puzzle, I've submitted $\dfrac{1}{\sin(x)cos(x)}$ (in another form) and I pointed out how by letting 1 = $sin^2x + cos^2x$ would lead to a solution. The formula can be rewritten as 1 = $(1 -cox^2x) + cos^2x$ for the numerator of the first fraction which of course leads to the same answer.

That method I refer to as the addition-subtraction method or ASM for short (which is applied to the numerators of fractions). This is a method that should be taught in calculus classes and demonstrated in texts, but unfortunately isn't. It's a powerful method for simplifying fractions and I'll now give you another example in the form of an easy puzzle.

Using ASM, transform $\dfrac{1}{x(1 + x^2)}$ into two fractions and then go on to integrate those two fractions.

I believe your idea is
$\displaystyle $\frac{1}{{x(1 + {x^2})}} = \frac{{(1 + {x^2}) - {x^2}}}{{x(1 + {x^2})}} = \frac{{1 + {x^2}}}{{x(1 + {x^2})}} - \frac{{{x^2}}}{{x(1 + {x^2})}} = \frac{1}{x} - \frac{x}{{1 + {x^2}}}$$

Hence
$\displaystyle $\int {\frac{{dx}}{{x(1 + {x^2})}}} = \int {(\frac{1}{x} - \frac{x}{{1 + {x^2}}}) \cdot dx} = \int {\frac{{dx}}{x}} - \int {\frac{{x \cdot dx}}{{1 + {x^2}}}} =$$
$\displaystyle $= \ln \left| x \right| - \frac{1}{2} \cdot \ln \left| {{x^2} + 1} \right| + C = \ln \left| {\frac{x}{{\sqrt {{x^2} + 1} }}} \right| + C$$
• Jan 14th 2011, 07:11 AM
Prove It
$\displaystyle \int{\frac{1}{\sin{x}\cos{x}}\,dx} = \int{\frac{\cos{x}}{\sin{x}\cos^2{x}}\,dx}$

$\displaystyle = \int{\frac{\cos{x}}{\sin{x}(1 - \sin^2{x})}\,dx}$

$\displaystyle = \int{\frac{\cos{x}}{\sin{x}(1 - \sin{x})(1 + \sin{x})}\,dx}$.

Now make the substitution $\displaystyle u = \sin{x}$ so that $\displaystyle \frac{du}{dx} = \cos{x}$ and the integral becomes

$\displaystyle \int{\frac{1}{u(1 - u)(1 + u)}\,\frac{du}{dx}\,dx}$

$\displaystyle = \int{\frac{1}{u(1 - u)(1 + u)}\,du}$

$\displaystyle = \int{\frac{1}{u} - \frac{1}{2(u - 1)} - \frac{1}{2(u + 1)}\,du}$

$\displaystyle = \ln{|u|} - \frac{1}{2}\ln{|u-1|} - \frac{1}{2}\ln{|u + 1|} + C$

$\displaystyle = \ln{|\sin{x}|} - \frac{1}{2}\ln{|\sin{x} - 1|} - \frac{1}{2}\ln{|\sin{x} + 1|} + C$.
• Jan 14th 2011, 07:48 AM
Pranas
Quote:

Originally Posted by Prove It
$\displaystyle \int{\frac{1}{\sin{x}\cos{x}}\,dx} = \int{\frac{\cos{x}}{\sin{x}\cos^2{x}}\,dx}$

$\displaystyle = \int{\frac{\cos{x}}{\sin{x}(1 - \sin^2{x})}\,dx}$

$\displaystyle = \int{\frac{\cos{x}}{\sin{x}(1 - \sin{x})(1 + \sin{x})}\,dx}$.

Now make the substitution $\displaystyle u = \sin{x}$ so that $\displaystyle \frac{du}{dx} = \cos{x}$ and the integral becomes

$\displaystyle \int{\frac{1}{u(1 - u)(1 + u)}\,\frac{du}{dx}\,dx}$

$\displaystyle = \int{\frac{1}{u(1 - u)(1 + u)}\,du}$

$\displaystyle = \int{\frac{1}{u} - \frac{1}{2(u - 1)} - \frac{1}{2(u + 1)}\,du}$

$\displaystyle = \ln{|u|} - \frac{1}{2}\ln{|u-1|} - \frac{1}{2}\ln{|u + 1|} + C$

$\displaystyle = \ln{|\sin{x}|} - \frac{1}{2}\ln{|\sin{x} - 1|} - \frac{1}{2}\ln{|\sin{x} + 1|} + C$.

$\displaystyle $\int {\frac{1}{{\sin x\cos x}}{\mkern 1mu} dx} = \int {\frac{{\cos x}}{{\sin x{{\cos }^2}x}}{\mkern 1mu} dx} = \int {\frac{{d(\tan x)}}{{\tan x}}{\mkern 1mu} } = \ln \left| {\tan x} \right| + C$$

Since
$\displaystyle $d(\tan x) = \frac{{dx}}{{{{\cos }^2}x}}$$
• Jan 14th 2011, 11:21 AM
wonderboy1953
Very good
Quote:

Originally Posted by Pranas
I believe your idea is
$\displaystyle $\frac{1}{{x(1 + {x^2})}} = \frac{{(1 + {x^2}) - {x^2}}}{{x(1 + {x^2})}} = \frac{{1 + {x^2}}}{{x(1 + {x^2})}} - \frac{{{x^2}}}{{x(1 + {x^2})}} = \frac{1}{x} - \frac{x}{{1 + {x^2}}}$$

Hence
$\displaystyle $\int {\frac{{dx}}{{x(1 + {x^2})}}} = \int {(\frac{1}{x} - \frac{x}{{1 + {x^2}}}) \cdot dx} = \int {\frac{{dx}}{x}} - \int {\frac{{x \cdot dx}}{{1 + {x^2}}}} =$$
$\displaystyle $= \ln \left| x \right| - \frac{1}{2} \cdot \ln \left| {{x^2} + 1} \right| + C = \ln \left| {\frac{x}{{\sqrt {{x^2} + 1} }}} \right| + C$$

Exactly what I had in mind. Normally this problem would be solved by the method of partial fractions which involves some algebra, but this way is much more efficient (wouldn't you agree?)
• Jan 14th 2011, 11:40 AM
Pranas
Quote:

Originally Posted by wonderboy1953
Exactly what I had in mind. Normally this problem would be solved by the method of partial fractions which involves some algebra, but this way is much more efficient (wouldn't you agree?)

Indeed, foreseeing the possibilities is always a good thing in mathematics