How do you evaluate #\int _ { 0} ^ { 1} ( x \sin ( x ) + \sin x ) d x#?

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Answer 1 · 2018-02-05T15:23:49

The answer is #=0.76#

Perform the indefinite integral by integration by parts

#intuv'=uv-intu'v#

Here,

#u=(x+1)#, #=>#, #u'=1#

#v'=sinx#, #=>#, #v=-cosx#

Therefore,

#int(x+1)sinxdx=-(x+1)cosx+intcosxdx#

#=-(x+1)cosx+sinx+C#

Then, the definite integral is

#int_0^1(x+1)sinxdx=[-(x+1)cosx+sinx]_0^1#

#=(-2cos1+sin1)-(-cos0+sin0)#

#=sin(1)-2cos(1)+1#

#=0.76#