Question

Найти производную функции: y=7Sin(6x^5+7x^3+3x^2) И Вычислить интеграл по частям: ∫(7x+9)Sin7xdx

Answer 1

$[7sin(6x^5+7x^3+3x^2)]'=\\ =7cos(6x^5+7x^3+3x^2)*(6x^5+7x^3+3x^2)'=\\ =7cos(6x^5+7x^3+3x^2)*(30x^4+21x^2+6x)$
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$\int\limits {(7x+9)sin7x} \, dx = \frac{1}{7} \int\limits {7x*sin(7x)} \, d(7x)+\frac{9}{7}\int\limits {sin(7x)} \, d(7x) =\\ =\frac{1}{7} \int\limits {t*sin(t)} \, dt+\frac{9}{7}\int\limits {sin(t)} \, dt =\\ =\frac{1}{7} \int\limits {t} \, d(-cos(t))-\frac{9}{7}cos(t)=\\ =-\frac{1}{7}[tcos(t)- \int\limits {cos(t)} \, dt ]-\frac{9}{7}cos(t)=\\ =\frac{1}{7}[tcos(t)-sin(t) ]-\frac{9}{7}cos(t)+C,\ where\ t=7x$