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Question : 77 of 160
Marks:
+1,
-0
Solution:
‌sin‌2x+cos‌2‌x∣dx‌⇒|sin‌2x+cos‌2‌x|=√2|sin‌(2x+‌)|‌⇒sin‌2x+cos‌2‌x∣dx‌=√2|sin‌(2x+‌)|‌dxLet
u=2x+‌‌du=2‌dx‌dx=‌du‌x=5π,u=10π+‌‌x=25π,u=50π+‌‌=√2∫10π+‌50π+‌|sin‌u|‌du‌=‌∫10π+‌50π+‌|sin‌u|duThe period of
|sin‌u| is
Ï€.
The length of the interval of integration is
‌=(50π+‌)−(10π+‌)=40π‌=‌⋅40π|sin‌u|du‌=‌×40×2‌=40√2
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