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Question : 74 of 160
Marks:
+1,
-0
Solution:
‌∫‌‌sin‌2x⋅cos2x(sin‌x+cos‌x)‌dx‌=sin‌3xcos2x‌dx‌+∫‌‌sin‌2xcos3x‌dx‌=∫‌‌sin‌x⋅(1−cos2x)cos2x‌dx‌+∫‌‌cos‌x(1−sin‌2x)⋅sin‌2x‌dx‌=∫‌‌sin‌x⋅cos2x⋅dx‌‌‌−∫‌‌sin‌x⋅cos4x‌dx‌+∫‌‌cos‌x⋅sin‌2x⋅dx−∫‌‌cos‌x⋅sin‌4x‌dx‌=[‌+‌+‌−‌]−‌π‌=(0+0+‌−‌)−(0+0−‌+‌)‌=‌−‌=‌=‌
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