Question

I want to know how to integrate sin(cosx) dx

Answer 1

\[\let u=\sin(cosx)\]\[\sin^{-1} (u)=cosx\]
\[x=\cos^{-1} (\sin^{-1} (u))\]\[x=\cos^{-1} (cosx)\]
from the graph;\[x=\sqrt{1-u^2}\]
\[dx=\frac{ u }{ (1-u^2)^\frac{ 3 }{ 2 } }du\]
\[\int\limits_{}^{}(\sin(cosx))dx=\int\limits_{}^{}(\frac{ u^2 }{ (1-u^2)^\frac{ 3 }{ 2 } })du\]
\[\let u=sinx, then du=cosxdx\]substitute in the equation then you will get
\[\int\limits_{}^{}(cosxdx/sinx)\]\[\let u=sinx. then du=cosxdx\]\[\int\limits_{}^{}(\frac{ 1 }{u })du=\ln(u)+c \].

Answer 2

Pynewn, your answer is ln(sinx)+C, I derivative ln(sinx)+c d/dx = cosx/sinx, it is the original function.

Answer 3

you need to use the expansion series to get this integral done

Answer 4

Ahmad can you plz do it using expansion series

Answer 5

if you know the how to use the series of expansion you will be having no trouble there, actually even I don't know much about them.
using the series at x=0
sin(1)-1/2x^2cos(1)+x^4/4!cos(1)-...
just integrate each term individually now
xsin(1)+.....