integrals. How do you solve integrals?

The indefinite integral of y^4 is (1/5)y^5, so the definite integral from 0 to 2 is (1/5)(2)^5 - (1/5)(0)^2 which is 32/5 - 0 which is just 32/5. I don't know where you get y^5, but the power rule states that for cy^n, the integral of cy^n is just (c/(n+1))y^(n+1). For definite integrals from a to b (for this particular case), the formula is just (c/(n+1))b^(n+1) - (c/(n+1))a^(n+1).

Don't know what's up with that y^5 up there, but here goes your integral...

int y^4 from 0 to 2 = y^5/5

2^5/5 - 0^5/5

32/5 - 0

integral = 32/5

I can't explain here how to solve integrals. There are many different methods for many different types of integrals. Yours is the simplest, but explaining why and how would take a long, long time.

You don't "solve" integrals. You solve equations. You evaluate integrals. i) Try writing cos^5(x) as cos(x) (1-sin^2(x)]^2. and try u = sin(x) ii) similar, but try u = sqrt(sin(x)) instead (if the first doesn't get you there)