Answer
$\int4x^7- 3cos(x)dx = \frac{1}{2}x^8 -3sin(x) + C$
Work Step by Step
Rules to remember:
$\int sin(x)dx = -cos(x)$, $\int cos(x)dx = sin(x)$
$sin(x)' = cos(x)$, $cos(x)' = -sin(x)$
Anti-differentiation: $\int4x^7- 3cos(x)dx = 4\times\frac{1}{8}x^8 -3sin(x) + C= \frac{1}{2}x^8 -3sin(x) + C$
Checking with differentiation:
$(\frac{1}{2}x^8 -3sin(x) + C)'= \frac{1}{2}\times8x^7 -3cos(x)= 4x^7 -3cos(x)$
