Question

Alguém por favor?
Calcule a integral por substituição:
∫6t a quarta sen(2t a quinta +3) dt

Answer 1

$\displaystyle\int6t^{4}sen(2t^{5}+3)dt$

Seja $a=2t^{5}+3$

Derivando 'a':

$da=10t^{4}dt~~~\therefore~~~\dfrac{6}{10}da=\dfrac{6}{10}10t^{4}dt~~~\therefore~~~\boxed{\boxed{6t^{4}dt=\dfrac{3}{5}da}}$

Então:

$\displaystyle\int6t^{4}sen(2t^{5}+3)dt=\int sen(2t^{5}+3)\cdot6t^{4}dt\\\\\\\int6t^{4}sen(2t^{5}+3)dt=\int sen(a)\cdot\dfrac{3}{5}da\\\\\\\int6t^{4}sen(2t^{5}+3)dt=\dfrac{3}{5}\int sen(a)da\\\\\\\int6t^{4}sen(2t^{5}+3)dt=\dfrac{3}{5}[-cos(a)]+constante\\\\\\\boxed{\boxed{\int6t^{4}sen(2t^{5}+3)dt=-\dfrac{3}{5}cos(2t^{5}+3)+constante}}$