Question

How to integrate
∫$e^{2lnsinx}$ + dx

Answer 1

Answer:

Step-by-step explanation:

We have,

$\sf{\displaystyle\int\,e^{\displaystyle2\ln(sin(x))}\,dx}$

$\sf{=\displaystyle\int\,e^{\displaystyle\ln(sin^2(x))}\,dx}$

$\sf{=\displaystyle\int\,sin^2(x)\,dx}$

$\sf{=\displaystyle\int\,\dfrac{1-cos(2x)}{2}\,dx}$

$\sf{=\displaystyle\int\,\dfrac{1}{2}\,dx-\int\dfrac{cos(2x)}{2}\,dx}$

$\sf{=\displaystyle\dfrac{1}{2}\int\,dx-\dfrac{1}{2}\int\,cos(2x)\,dx}$

$\sf{=\displaystyle\dfrac{1}{2}x-\dfrac{1}{2}\cdot\dfrac{sin(2x)}{2}+C}$

$\sf{=\displaystyle\dfrac{x}{2}-\dfrac{sin(2x)}{4}+C}$