Calculus Homework?

Question

I have a 30 problem assignment due tonite and these are the last few questions I cannot get. I just went back to school to change my major and haven't had calc a in 4 years so don't make fun of me if these are extremely simple please =)

[INT FROM 0 to 1] x^(2)((fifth route)e^(x)) dx

[INT FROM 1 to 6] sq route(t) lntdt

[INT] 340cos^(4)(20x)dx

[INT] sin(4x)cos(13x) dx

[INT FROM 0 to Pi/2] sin^(5) x cos^(16) xdx

Any help is super appreciated. Thanks y'all!

Hi TheMathStudent,
Thanks for the help. I'm not sure if I understand what you did for the second one though? And also, my last question is correct. It would read:

"evaluate the definate integral from 0 to pi/2 of sin to the fifth power x times cosine to the sixteenth power x dx."

thanks again!

Answer 1

1) M=int(x^2 * e^(x/5) dx )
=x^2 *5e^(x/5)-int(2x* 5e^(x/5) dx)
=x^2 *5e^(x/5)-10[x*5e^(x/5)-int(1*5e^(x/5) dx )
=x^2 *5e^(x/5)-50x*e^(x/5)+50*5e^(x/5) +c
now apply the limits.
2)int(ln(t)t^(1/2) dt
=ln(t)*(1/2)t^(-1/2)-int( (1/t)(1/2)t^(-1/2) dt )
3)cos^(2)x=(1+cos(2x))/2
(cos^(2)20x)^2=[(1+cos(40)x)/2]^2
now expand do the integration.
4)cosAsinB=(1/2)[sin(A+B)-sin(A-B)]
here (1/2)[sin(17)x -sin(9)x]
when we integrate ,
(1/2)[-cos(17x) /17 +cos(9x) /9 ]
5) iam not so sure about the last one.can you check whether your question is right.

Answer 2

I'm not going to give you the complete solutions, as I don't like doing other people's homework for them. But here are some hints to get you started:

#1: Write as ∫x²e^(x/5) dx. Apply integration by parts twice.
#2: It's square _root_, not square route. And again, apply integration by parts once (differentiate ln t, and remember that √x=x^(1/2), so you can integrate using the power rule)
#3: Apply the identity cos² x = (1+cos(2x))/2 twice
#4: Use the identity sin θ cos φ = (sin (θ+φ) + sin (θ-φ))/2. Alternatively, use the identities cos θ=(e^(iθ)+e^(-iθ))/2 and sin θ=(e^(iθ)-e^(-iθ))/(2i), which I find easier to remember and from which the former identity is easily derived.
#5: Note that sin^5 x cos^16 x = cos^16 x (1-cos² x)² sin x = (cos^20 x - 2 cos^18 x + cos^16 x) sin x, so just make the substitution u=cos x.

Answer 3

I don't have a clue, but I don't know how people do calculus! It looks so hard!! :O

I wish I could help you though, i'm just a junior in high school taking algebra 2 lol!