How do I evaluate the integral of sin(cos(x)) from 0 to pi/2? Thanks for the help.?

Question

I thought you could take the arcsin of both y and sin(cos(x)) and manipulate it around. Doesn't seem to work tho.

Answer 1

I don't think this can be solved in terms of elementary functions. Integration by parts, u-substitution, trigonometric identities, trigonometric substitution, partial fractions, and all of the Calculus 2 methods won't work.

Unless somebody proves me wrong.

Answer 2

Hi,

Thank goodness for TI83s. Type in y1 = sin(cos(x)) and set your window to xmin = 0, xmax = 2, ymin = -.2, ymax = .2. Go to the CALC commands and select #7, the integral. Type in the lower limit of 0, the upper limit of Pi/2, using the Pi button and voila! You get the area under the curve is .02741074. Not the biggest value, but there it is.

I hope that helps. Have a good day.

Answer 3

Are you sure you wrote the question correctly? As written, Puggy is right, it can't be solved by basic calculus.

This question would be a lot easier if it was:

integral of sin(cos(x))(-sinx) from 0 to pi/2

In that case, the answer would be:
cos(1) - 1

But since that's not your question, I don't have an answer for you.

Answer 4

The integral is equal to the number
1/2*Pi*StruveH[0,1]
Where StruveH[0,1] stands for
Struve function which satisfies
the differential equation
y''+y'+y=2/Pi