Integral Exercise asked by Katrine Linda and the idea of solution credit to Ngô Nghiệp

Exercise:

Integrate, $\int{\frac{dx}{\sin \left( x+2 \right)\sin \left( x+3 \right)}}$

Solution: we have $\sin \left( x+3-x-2 \right)=\sin \left( \left( x+3 \right)-\left( x+2 \right) \right)=\sin \left( 1 \right)$

But $\sin \left( \left( x+3 \right)-\left( x+2 \right) \right)=\cos \left( x+2 \right)\sin \left( x+3 \right)-\cos \left( x+3 \right)\sin \left( x+2 \right)$

Notice that $\frac{1}{\sin \left( x+2 \right)\sin \left( x+3 \right)}=\frac{\sin \left( 1 \right)}{\sin \left( 1 \right)}\times \frac{1}{\sin \left( x+2 \right)\sin \left( x+3 \right)}$

Hence $\frac{1}{\sin \left( 1 \right)}\times \frac{\sin \left[ \left( x+3 \right)-\left( x+2 \right) \right]}{\sin \left( x+2 \right)\sin \left( x+3 \right)}=\frac{1}{\sin \left( 1 \right)}\left[ \frac{\cos \left( x+2 \right)}{\sin \left( x+2 \right)}-\frac{\cos \left( x+3 \right)}{\sin \left( x+3 \right)} \right]$

So $\int{\frac{dx}{\sin \left( x+2 \right)\sin \left( x+3 \right)}=\frac{1}{\sin \left( 1 \right)}\int{\frac{\cos \left( x+2 \right)}{\sin \left( x+2 \right)}dx-\frac{1}{\sin \left( 1 \right)}\int{\frac{\cos \left( x+3 \right)}{\sin \left( x+3 \right)}dx}}}$

$=\frac{1}{\sin \left( 1 \right)}\left[ \int{\frac{d\left( \sin \left( x+2 \right) \right)}{\sin \left( x+2 \right)}}-\int{\frac{d\left( \sin \left( x+3 \right) \right)}{\sin \left( x+3 \right)}} \right]=\frac{1}{\sin \left( 1 \right)}\left[ \ln \left| \sin \left( x+2 \right) \right|-\ln \left| \sin \left( x+3 \right) \right| \right]+c$

$=\csc \left( 1 \right)\ln \left| \frac{\sin \left( x+2 \right)}{\sin \left( x+3 \right)} \right|+c$