Evaluate the definite integral.?

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Answer 1 · 2018-05-28T18:10:32

#1/6*(1-cos(1))#

Sunstituting
#t=sin(6*x)# then we get
#1/6dt=cos(6x)dx# and we get

#1/6int_0^1 sin(t)dt=1/6*(1-cos(1))#

Answer 2 · 2018-05-28T18:18:15

I tried this:

Have a look:
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Answer 3 · 2018-05-28T18:18:25

# 1/6(1-cos1)#.

Suppose that, #I=int_0^(pi/12) cos6xsin(sin6x)dx#.

Subst. #sin6x=y. :. 6cos6xdx=dy#.

Also, when #x=0,y=sin(6xx0)=sin0=0# and

when #x=pi/12, y=sin(6*pi/12)=sin(pi/2)=1#.

#:. I=int_0^(pi/12) cos6xsin(sin6x)dx#,

#=1/6int_0^(pi/12) (sin(sin6x))(6cos6x)dx#,

#=1/6int_0^1 sinydy#,

#1/6[-cosy]_0^1#,

#=-1/6[cos1-cos0]#.

# rArr I=1/6(1-cos1)#.