# Math Help - The addition-subtraction puzzle (calculus)

1. ## 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.

2. 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.
$\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}}}} =$$
$\displaystyle $= \ln \left| x \right| - \frac{1}{2} \cdot \ln \left| {{x^2} + 1} \right| + C = \ln \left| {\frac{x}{{\sqrt {{x^2} + 1} }}} \right| + C$$

3. $\displaystyle \int{\frac{1}{\sin{x}\cos{x}}\,dx} = \int{\frac{\cos{x}}{\sin{x}\cos^2{x}}\,dx}$

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

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

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

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

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

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

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

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

4. 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}}$$

5. ## Very good

Originally Posted by Pranas
$\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}}}$$
$\displaystyle $\int {\frac{{dx}}{{x(1 + {x^2})}}} = \int {(\frac{1}{x} - \frac{x}{{1 + {x^2}}}) \cdot dx} = \int {\frac{{dx}}{x}} - \int {\frac{{x \cdot dx}}{{1 + {x^2}}}} =$$
$\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$$