1. 1/2 cos(x) sin(x) + x/2 + C

1. Using formula 8. from the notes, with n= 2, tells us that ∫cos²(x) dx = 1/2 cos(x) sin(x) + 1/2 ∫ cos⁰(x) dx Since cos⁰(x) = 1, the integral on the right side evaluates to: ∫cos⁰(x) dx = ∫dx= x + c We can put that back in our original integral, and get: ∫cos²(x) dx = 1/2 cos(x) sin(x) + x/2 + C The answer given by formula 2 is: ∫cos²(x) dx = x/2 + 1/4 sin(2x) + C If we use the identity sin(2x) = 2 sin(x) cos (x), we see that the results are the same.

1. 1/4 cos³(x) sin(x) + 3/8 cos(x) sin(x) + 3/8 x + C

1. We can use formula 8. from the notes with n = 4, to reduce the power of cos(x) in the expression: ∫cos⁴x dx = 1/4 cos³(x) sin(x) + 3/4 ∫cos²(x) dx To evaluate the integral on the right side, we can use our answer to the previous problem (or, alternatively, formula 2. in the notes). Our answer was ∫cos²(x) dx = 1/2 cos(x) sin(x) + x/2 + C Putting that in, we get: ∫cos⁴x dx = 1/4 cos³(x) sin(x) + 3/8 cos(x) sin(x) + 3/8 x + C

1. Arcsin [ln(x)] + c

1. This looks like formula 30. from the notes, except we have ln²(x) under the square root instead of x². Let’s try using substitution first, with u = ln²(x). Then du = 1/x dx, and we get ∫1/x [1 – ln²(x)]^-1/2 dx = ∫[1 – u²]^-1/2 du The factor 1/x disappeared, and we’re in the territory of formula 28. Using it gives us ∫1/x [1 – ln²(x)]^-1/2 dx = ∫[1 – u²=>]^-1/2 du = arcsin(u) + c = arcsin [ln(x)] + c

1. ± x + C

1. This looks very much like formula 28. in the notes, except it’s for sin(x) instead of x. But we have cos(x) up front, which is the derivative of sin(x): perfect case for substitution u = sin(x), du = cos(x) dx, followed by formula 28. for u: ∫cos(x) [1 – sin²(x)]^-1/2 dx = ∫[1 – u²]^-1/2du = arcsin(u) + c = arcsin [sin(x)] + c This is interesting. We are getting arcsin[sin(x)] + c = ± x + nπ + c The nπ + c can be replaced by a new constant C, and we get, finally, that ∫cos(x) [1 – sin²(x)]^-1/2 dx = ± x + C That seems way too easy for such a complicated expression. We could have simplified it right at the beginning, by noting that 1 – sin²(x) = cos²(x). The integral is ∫cos(x) [1 – sin²(x)]^-1/2 dx = ∫cos(x)/| cos(x) | dx= ∫± dx = ± x + C Even with tables at hand, it pays to check for simplifications first, before looking anything up.

1. -x/2 [1 – x²]^1/2 + 1 /2 arcsin(x) + c

1. What to take for dv? We know how to integrate both x and x², but setting either to dv will make the situation worse, as we have seen before in similar problems: the power of x will increase, and that’s not helpful in this case. With the hint to use formula 28., we may as well set dv = [1 – x²]^-1/2, which gives us the variables as follows: The easiest factor to integrate is dv = x[1 – x²]-1/2, using substitution with u= x²: ∫x[1 – x²]^-1/2 = -1/2 [1 – x²]^1/2 + C With this, our substitution variables are: dv = x[1 – x2]-1/2, u = x, so du = 1, v = -[1 – x2]1/2 Integrating by parts, we get: ∫x²[1 – x²]-1/2 dx = – x [1 – x2]1/2 + ∫[1 – x2]1/2 dx We can use formula 24. for the right hand side integral: ∫[1 – x²]^1/2 dx = + x / 2 [1 – x²]^1/2 + 1 /2 arcsin(x) + c Putting this into our original answer gives: ∫x² [1 – x²]^-1/2 dx = -x [1 – x²]^1/2 + x/2 [1 – x²]^1/2 + 1 /2 arcsin(x) + c = -x / 2 [1 – x²]^1/2 + 1 /2 arcsin(x) + c This is exactly the same as formula 29.