Integral Exercise asked in Csb.go.lb in 1995

Exercise:

Integrate, $\int{\frac{dt}{t+2t\cos x+1}}$ where $t\in \mathbb{R}$

Solution: we have $\int{\frac{dt}{t+2t\cos x+1}=\int{\frac{dt}{t\left( 1+2\cos x \right)+1}}}$

If $1+2\cos x=0\Leftrightarrow \int{\frac{dt}{0t+1}=\int{dt=t+c}}$

If $1+2\cos x\ne 0$ Take $u=t\left( 1+2\cos x \right)+1\Leftrightarrow du=\left( 1+2\cos x \right)dt\Leftrightarrow dt=\frac{du}{1+2\cos x}$

So $\int{\frac{dt}{t\left( 1+2\cos x \right)+1}=\int{\frac{\frac{du}{1+2\cos x}}{u}}=\int{\frac{du}{u\left( 1+2\cos x \right)}=\frac{1}{1+2\cos x}\ln \left| u \right|+c}}$

                              $=\frac{1}{1+2\cos x}\ln \left| t\left( 1+2\cos x \right)+1 \right|+c$