Question

Evaluate the integral.

∫ 0 to pi/2 (5sin^(3/2)x * cosx * dx)

Answer 1

i dont think you can solve that integral with out a calculator...
if it was x^2 you could use a u substitution, or if it was sinxcosx you could use an identity. Are you allowed to use a graph?

Answer 2

here is a double angle if you want to try and play with this lol
sin(2x) = 2sin(x)cos(x)

Answer 3

I'm using substitution. Also this assignment cannot use graphs nor a calculator. Funny because I slowly creeped my hand towards my calculator and I had to pull back begrudgingly.

Answer 4

So far I have u=sinx and du=cosx*dx. So I get.

\[5\int\limits_{0}^{1}u^{3/2}*du\]

I got 1 and 0 from in integral from u(pi/2) and u(0) from the original integral.

Answer 5

Do I have to isolate dx from du=cosx*dx?

Answer 6

I must of miss read the formula. you can just integrate u^3/2 now which is (2/5)u^5/2
now just switch u back to sinx and plug in the upper minus the lower... Did that answer your question?

Answer 7

Hmmm. I checked I got .259811 as the answer. The answer is supposed to be 2 from what I've seen. I know the answer. I just need to understand the process. Sorry man. It didn't help. Maybe there was a small mistake?

Answer 8

note that you have a **sin to a power** with a **cos** conveniently there for when you apply the chain rule. that is the key.

is such a case start by thinking what could have differentiated to sin ^(3/2)x ? that 's sin(5/2) x, so now do the full differentiation.

d/dx(sin^(5/2) x = (5/2) • sin^ (3/2) x • cos x

that is almost what you are being asked to do save you have a factor of 1/2 to remove. so the integral is :

2sin^(5/2) x

between 0 and π/2 you get ......

Answer 9

\[5\int\limits_{}^{}(\cos x)(\sin x)^{3/2}~dx\]a simple u substitution will suffice. u=sin(x)
du = cos x dx

Answer 10

whether

(1) to change limits of integration, but then NOT substitute the original variable (the x) back into the antiderivative,

or

(2) to NOT CHANGE LIMITS of integration, but then to YES SUBSTITUTE the original variable back into the antiderivative


is entirely up to you (!)

Answer 11

I wonder why the hassle about this integral while it is only for u=sinx?