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**Ackbeet** Just looking at the integrand:

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

We can postulate, then, that

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

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

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

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

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

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

Integrating yields

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