Evaluate: integration[sin(x^0)dx]?

1 Answer
Yahia M.
Apr 7, 2018

#(sin1)*x+C#

or

#-pi/180cos(180/pix)+C#

Explanation:

First Answer
if You mean #x# to the power of zero:
First, #sinx^0=sin1#
so it's integration will be easy with the power rule:
#intsin1dx#=#(sin1)*x+C#
.

Second Answer
If you mean #x# degree then this will be the answer:
#x_(degree)#=#180/pix_(radian)#

and by substituting you get
#intsin(x^0)*dx=intsin(180/pix)*dx#

=#-cos(180/pix)/(180/pi)+C#

=#-pi/180cos(180/pix)+C#

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