Question

Como resolvo essa integral indefinida?

$\int ( \sqrt 2 seny - e^y/2) dx$

Answer 1

$\bullet\displaystyle\int\bigg[\sum\limits_{i=1}^{n}f_{i}(x)\bigg]\,dx=\sum\limits_{i=1}^{n}\int f_{i}(x)\,dx$

(vulgo "a integral da soma é a soma das integrais")

Dessa propriedade, tiramos que

$\bullet\displaystyle\int cf(x)\,dx=c\int f(x)\,dx$

(constante multiplicando o integrando sai da integral)
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$\displaystyle\int\bigg[\sqrt{2}sen(y)-\dfrac{~e^{y}}{2}\bigg]\,dy=\int\sqrt{2}sen(y)\,dy-\int\dfrac{~e^{y}}{2}\,dy\\\\\\\int\bigg[\sqrt{2}sen(y)-\dfrac{~e^{y}}{2}\bigg]\,dy=\sqrt{2}\int sen(y)\,dy-\int\dfrac{1}{2}e^{y}\,dy\\\\\\\int\bigg[\sqrt{2}sen(y)-\dfrac{~e^{y}}{2}\bigg]\,dy=\sqrt{2}\int sen(y)\,dy-\dfrac{1}{2}\int e^{y}\,dy$

Sabendo que

$\dfrac{d}{dy}-cos(y)=sen(y)~~~e~~~\dfrac{d}{dy}e^{y}=e^{y}$

temos que

$\displaystyle\int sen(y)\,dy=-cos(y)+C_{1}~~~e~~~\int e^{y}\,dy=e^{y}+C_{2}$

Logo:

$\displaystyle\int\bigg[\sqrt{2}sen(y)-\dfrac{~e^{y}}{2}\bigg]\,dy=\sqrt{2}[-cos(y)]-\dfrac{1}{2}e^{y}+constante\\\\\\\boxed{\boxed{\int\bigg[\sqrt{2}sen(y)-\dfrac{~e^{y}}{2}\bigg]\,dy=-\sqrt{2}cos(y)-\dfrac{e^{y}}{2}+constante}}$