# Easy calculus puzzle

• Dec 28th 2010, 01:07 PM
wonderboy1953
Easy calculus puzzle
One of the methods that's normally used is the method of partial fractions to convert complex fractions into easier ones.

Yet in the following I've never seen this method applied to trig functions where it's also effective. So for example can you solve by partial fractions the following?:

http://www.mathhelpforum.com/math-he...9a7b2f6e1b.png

If you can do the integration (which is solvable by other means, but the focus is on the method of partial fractions), then can you integrate the following also using the method of partial fractions?:

http://www.mathhelpforum.com/math-he...55bbec4058.png

• Dec 28th 2010, 01:51 PM
Ackbeet
Just looking at the integrand:

$\sec(x)\csc(x)=\dfrac{1}{\sin(x)}\,\dfrac{1}{\cos( x)}=\dfrac{A}{\cos(x)}+\dfrac{B}{\sin(x)}=\dfrac{A \sin(x)+B\cos(x)}{\cos(x)\sin(x)}.$

We can postulate, then, that

$\sec(x)\csc(x)=\dfrac{\sin(x)}{\cos(x)}+\dfrac{\co s(x)}{\sin(x)}.$

That is, $A=\sin(x), B=\cos(x).$

Integrating gives you the logarithm of the trig functions (one has a minus sign), since the derivatives of the denominators are like the numerators.

For the second problem, you have

$\dfrac{1}{\sin^{2}(x)}\,\dfrac{1}{\cos^{2}(x)}=\df rac{A}{\sin^{2}(x)}+\dfrac{B}{\cos^{2}(x)}=\dfrac{ A\cos^{2}(x)+B\sin^{2}(x)}{\sin^{2}(x)\cos^{2}(x)} .$

$A=B=1$ renders this an equality. Hence, we get

$\dfrac{1}{\sin^{2}(x)}\,\dfrac{1}{\cos^{2}(x)}=\df rac{1}{\sin^{2}(x)}+\dfrac{1}{\cos^{2}(x)}.$

Integrating yields

$\displaystyle\int\left[\dfrac{1}{\sin^{2}(x)}+\dfrac{1}{\cos^{2}(x)}\righ t]dx=-\cot(x)+\tan(x)+C.$
• Dec 28th 2010, 01:57 PM
wonderboy1953
Quote:

Originally Posted by Ackbeet
Just looking at the integrand:

$\sec(x)\csc(x)=\dfrac{1}{\sin(x)}\,\dfrac{1}{\cos( x)}=\dfrac{A}{\cos(x)}+\dfrac{B}{\sin(x)}=\dfrac{A \sin(x)+B\cos(x)}{\cos(x)\sin(x)}.$

We can postulate, then, that

$\sec(x)\csc(x)=\dfrac{\sin(x)}{\cos(x)}+\dfrac{\co s(x)}{\sin(x)}.$

That is, $A=\sin(x), B=\cos(x).$

Integrating gives you the logarithm of the trig functions (one has a minus sign), since the derivatives of the denominators are like the numerators.

For the second problem, you have

$\dfrac{1}{\sin^{2}(x)}\,\dfrac{1}{\cos^{2}(x)}=\df rac{A}{\sin^{2}(x)}+\dfrac{B}{\cos^{2}(x)}=\dfrac{ A\cos^{2}(x)+B\sin^{2}(x)}{\sin^{2}(x)\cos^{2}(x)} .$

$A=B=1$ renders this an equality. Hence, we get

$\dfrac{1}{\sin^{2}(x)}\,\dfrac{1}{\cos^{2}(x)}=\df rac{1}{\sin^{2}(x)}+\dfrac{1}{\cos^{2}(x)}.$

Integrating yields

$\displaystyle\int\left[\dfrac{1}{\sin^{2}(x)}+\dfrac{1}{\cos^{2}(x)}\righ t]dx=-\cot(x)+\tan(x)+C.$

Excellent. What I was looking for. It's interesting that Schaum's outline and my math textbook didn't have these examples.
• Dec 28th 2010, 07:03 PM

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

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

if $cos(x) = t$

$sinxdx=-dt$

$\int \frac{-1}{(1-t^2)t}dt$

$\int \frac{A}{t-1}+\frac{B}{t+1}+\frac{C}{t} dt$
• Dec 29th 2010, 03:15 PM
wonderboy1953
Quote:

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

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

if $cos(x) = t$

$sinxdx=-dt$

$\int \frac{-1}{(1-t^2)t}dt$

$\int \frac{A}{t-1}+\frac{B}{t+1}+\frac{C}{t} dt$

Ackbeet's is simpler since he's only has an A and B in his partial fractions method while you have A, B and C (however check out my next puzzle which directly relates to what you just posted).
• Jan 11th 2011, 09:30 AM
wonderboy1953
A simpler method for the two problems
From the well-known trig formula, $sin^2x + cos^2x = 1$, you can substitute for 1 in the numerator to break down the fractions into simpler ones in the two problems I gave in the first post on this thread.