Integral exercises Level 2

Exercise:

Integrate, $\int{\sqrt{1+{{e}^{x}}}dx}$

Solution: Let ${{u}^{2}}=1+{{e}^{x}}\Leftrightarrow 2udu={{e}^{x}}dx\Leftrightarrow dx=\frac{2u}{{{u}^{2}}-1}du$

So $\int{\sqrt{1+{{e}^{x}}}dx=\int{\frac{2{{u}^{2}}}{{{u}^{2}}-1}du=2\int{\frac{{{u}^{2}}-1+1}{{{u}^{2}}-1}du=2\int{du+2\int{\frac{du}{{{u}^{2}}-1}}}}}}$

\[=2\sqrt{1+{{e}^{x}}}-2\int{\frac{du}{1-{{u}^{2}}}=2\sqrt{1+{{e}^{x}}}-2\operatorname{arctanh}\left( u \right)+c=2\sqrt{1+{{e}^{x}}}-2\operatorname{arctanh}\sqrt{1+{{e}^{x}}}+c}\]


Exercise:

Compute, $\int_{\ln \frac{\pi }{6}}^{\ln \frac{\pi }{3}}{3{{e}^{x}}\cos \left( 3{{e}^{x}} \right)dx}$

Solution: Let $u=3{{e}^{x}}\Leftrightarrow du=3{{e}^{x}}dx$ so $u\left( \ln \frac{\pi }{3} \right)=3{{e}^{\ln \frac{\pi }{3}}}=3\left( \frac{\pi }{3} \right)=\pi $ & $u\left( \ln \frac{\pi }{6} \right)=3\left( \frac{\pi }{6} \right)=\frac{\pi }{2}$

So $\int_{\ln \frac{\pi }{6}}^{\ln \frac{\pi }{3}}{3{{e}^{x}}\cos \left( 3{{e}^{x}} \right)dx}=\int_{u\left( \ln \frac{\pi }{6} \right)}^{u\left( \ln \frac{\pi }{3} \right)}{\cos u\,du}=\left[ \sin u \right]_{\frac{\pi }{2}}^{\pi }=\sin \pi -\sin \frac{\pi }{2}=-1$