Question

Evaluate:$\int \frac{1+sin2x}{1+cos2x}{e}^{2x}dx$

Answer 1

Put $2x=t\Rightarrow dx=\frac{1}{2}dt\therefore \int \frac{1+sin2x}{1+cos2x}{e}^{2x}dx=\frac{1}{2}\int \frac{1+sin2x}{1+cos2x}{e}^{t}dt$
$\Rightarrow =\frac{1}{2}\int (\frac{1}{2co{s}^2\left(\frac{t}{2}\right)}+\frac{2sin\left(\frac{t}{2}\right)cos\left(\frac{t}{2}\right)}{2co{s}^2\left(\frac{t}{2}\right)})e^{t}dt$
$\Rightarrow =\frac{1}{2}\int (\frac{se{c}^2\left(\frac{t}{2}\right)}{2}+tan\left(\frac{t}{2}\right))e^{t}dt$
$\because f\left(t\right)=tan\left(\frac{t}{2}\right),f^{\prime }\left(t\right)=\frac{se{c}^2\left(\frac{t}{2}\right)}{2}\therefore \text{using}\int \left[f\right(t)+f^{\prime }(t\left)\right]e^{t}dt=f\left(t\right)e^t+C$
$\therefore \int \frac{1+sin2x}{1+cos2x}{e}^{2x}dx=\frac{1}{2}{e}^{t}tan\left(\frac{t}{2}\right)+C=\frac{1}{2}{e}^{2x}tanx+C$