Evaluate the integral absolute value of \sin(x+\pi ) from 0 to

Edmund Adams

Edmund Adams

Answered question

2021-12-07

Evaluate the integral
absolute value of sin(x+π) from 0 to 3π2

Answer & Explanation

menerkupvd

menerkupvd

Beginner2021-12-08Added 12 answers

Step 1
To evaluate the integral
03π2|sin(x+π)|dx
Step 2
03π2|sin(x+π)|dx
=03π2|sinxcosπ+cosxsinπ|dx (using sin(a+b)=sinacosb+cosasinb)
=03π2|sinx(1)+cosx(0)|dx
=03π2|sinx|dx
=03π2sinxdx
=[cosx]03π2
=[cos(3π2)(cos0)]
=[-0]-[-1]
=1
Hence, 03π2|sin(x+π)|dx=1.

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