A square root is equal to 1/2...
=t^4(1/2)+t^2(1/2)+4^(1/2)......the square root is removed
=t^2+t+2
g'(t)=2t+1
Hi, I have 3 problems here need to be solved, help me please
1.Evaluate ∫ sin2x sec^5(2x)dx
2. If f(x)=√x and P is a partition of [1,16] determined by 1,3,5,7,9,16,find a Riemann sum Rp of f by choosing the numbers 1,4,5,9 and 9(yes 9 again) in the subintervals of P
3. Given G(x)=∫(limit 1/x to x) √t^4+t^2+4 dt, find G'(x)
2 answers
sin u sec^5 u
= tan u sec^4 u
= sec u * tan u * sec^3 u du
if v = sec u, then that is v^3 dv
So, you have
∫ sin2x sec^5(2x)dx
1/2 ∫ sin2x sec^5(2x) 2 dx
= 1/2 ∫ sec^3(2x) d(sec(2x))
= 1/8 sec^4(2x) + C
There are lots of good Riemann sum calculators online.
Using the 2nd Fundamental Theorem of Calculus, G'(x) =
[√(x^4+x^2+4) * (1)] - [√(1/x^4 + 1/x^2 + 4) * (-1/x^2)]
= √(x^4+x^2+4) + √(4x^4+x^2+1)/x^4
= tan u sec^4 u
= sec u * tan u * sec^3 u du
if v = sec u, then that is v^3 dv
So, you have
∫ sin2x sec^5(2x)dx
1/2 ∫ sin2x sec^5(2x) 2 dx
= 1/2 ∫ sec^3(2x) d(sec(2x))
= 1/8 sec^4(2x) + C
There are lots of good Riemann sum calculators online.
Using the 2nd Fundamental Theorem of Calculus, G'(x) =
[√(x^4+x^2+4) * (1)] - [√(1/x^4 + 1/x^2 + 4) * (-1/x^2)]
= √(x^4+x^2+4) + √(4x^4+x^2+1)/x^4