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Joint Entrance Examination
Mathematics
Indefinite Integrals
Question 159
Question 159 of 416
Easy
NUMERICAL
The value of the integral
∫
0
π
∣
sin
2
x
∣
d
x
\displaystyle\int_0^\pi|\sin2x|\,dx
∫
0
π
∣
sin
2
x
∣
d
x
is .......... Answer: 2
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Official solution
By periodicity,
∫
0
π
∣
sin
2
x
∣
d
x
=
2
∫
0
π
/
2
sin
2
x
d
x
=
2
[
−
cos
2
x
2
]
0
π
/
2
=
−
[
cos
π
−
cos
0
]
=
2
\displaystyle\int_0^\pi|\sin2x|\,dx = 2\int_0^{\pi/2}\sin2x\,dx = 2\left[-\frac{\cos2x}2\right]_0^{\pi/2} = -\big[\cos\pi-\cos0\big] = 2
∫
0
π
∣
sin
2
x
∣
d
x
=
2
∫
0
π
/2
sin
2
x
d
x
=
2
[
−
2
cos
2
x
]
0
π
/2
=
−
[
cos
π
−
cos
0
]
=
2
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