Question

$\color{green}{\int_{0}^2 \begin{gathered}\rm \sqrt{2}\bigg(\dfrac{1}{ \sqrt{2} } cosx + \frac{1}{ \sqrt{2} } sinx\bigg) \\ \end{gathered} \rm dx}$​

Answer 1

$\displaystyle \sf\int_{0}^{2} \: \sqrt{2} \bigg( \dfrac{1}{ \sqrt{2} } \cos(x) + \frac{1}{ \sqrt{2} } \sin(x) \bigg)dx$

$\displaystyle \sf \int \: \sqrt{2} \bigg( \frac{1}{ \sqrt{2}} \times \cos(x) + \frac{1}{ \sqrt{2} } \times \sin(x) \bigg) \: dx$

$\displaystyle \sf \sqrt{2} \times \int \frac{1}{ \sqrt{2} } \times \cos(x) + \frac{1}{ \sqrt{2} } \times \sin(x) \: dx$

$\displaystyle \sf \sqrt{ 2} \times \int \: \frac{ \cos(x) }{ \sqrt{2} } + \frac{ \sin(x) }{ \sqrt{2} } \: dx$

Rationalize the denominator, by multiplying √2.

$\displaystyle \sf \sqrt{2} \times \int \: \frac{ \green{ \sqrt{2} } \cos(x) }{2} + \frac{ \green{\sqrt{2} } \sin(x) }{2} \: dx$

$\displaystyle \sf \: \sqrt{2} \times \int \frac{ {2}^{ \frac{1}{2} } \times \cos(x) }{2} + \frac{ {2}^{ \frac{1}{2} } \times \sin(x) }{2} \: dx$

$\displaystyle \sf \: \sqrt{2} \times \int \frac{ \cos(x) }{ {2}^{ \frac{1}{2} } } + \frac{ \sin(x) }{ {2}^{ \frac{1}{2} } } \: dx$

$\displaystyle \sf \sqrt{2} \times \int \: \frac{ \cos(x) + \sin(x) }{ {2}^{ \frac{1}{2} } } \: dx$

$\displaystyle \sf 1 \times \int \: \cos(x) + \sin(x) \: dx$

$\displaystyle \sf \int \: \cos(x) + \sin(x) \: dx$

$\displaystyle \sf \int \: \cos(x) \: dx \: + \int \: \sin(x) \: dx$

$\displaystyle \sf \: \sin(x) - \cos(x)$

$\sf \bigg( \sin(x) - \cos(x) \bigg) \bigg|_{0}^{2}$

$\sf \: \sin(2) - \cos(2) - \bigg( \sin(0) - \cos(0) \bigg)$

$\green{ \sf \: \sin(2) - \cos(2) + 1}$