Question

Evaluate the following: $\int e^{2x}\left(\frac{1+sin2x}{1+cos2x}\right)dx$

Answer 1

$I=\int e^{2x}\left(\frac{1-1sin2x}{1-1cos2x}\right)dx$

$=\int e^{2x}(\frac{1}{1+1cos2x}+\frac{sin2x}{1+cos2x})dx$

$=\int e^{2x}(\frac{1}{2co{s}^{2}x}+\frac{2sinxcosx}{2co{s}^{2}x})dx$

$=\int e^{2x}(\frac{1}{2}se{c}^{2}x+tanx)dx$

Put $2x=t$

$\Rightarrow dx=\frac{1}{2}dt$

$=\int e^t(\frac{1}{2}se{c}^2\frac{t}{2}+tan\frac{t}{2}).\frac{1}{2}dt$

$=\frac{1}{2}\int e^t(tan\frac{t}{2}+\frac{1}{2}se{c}^2\frac{t}{2})dt$

We know $\int e^x\left(l\right(x)+l^1(x\left)\right)dx=e^{x}l\left(x\right)+C$

$=\frac{1}{2}{e}^{t}tan\frac{t}{2}+C$

$I=\frac{1}{2}{e}^{t}tan\frac{t}{2}+C$