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Question : 62 of 115
Marks:
+1,
-0
Solution:
‌‌| x2sin‌x⋅cos‌x |
| sin‌4x+cos4x |
‌dx ‌=‌| sin‌x⋅cos‌x |
| sin‌4x+cos4x |
(x2−(π−x)2)‌dx ‌=‌| sin‌x⋅cos‌x(2πx−π2) |
| sin‌4x+cos4x |
‌=2π‌‌| xsin‌x‌cos‌x |
| sin‌4x+cos‌4‌x |
‌dx −π2‌‌| sin‌x‌cos‌x |
| sin‌4x+cos‌4‌x |
‌dx ‌=2π⋅‌‌‌| sin‌xcos4x |
| sin‌4x+cos4x |
‌dx −π2‌‌| sin‌xcos4x |
| sin‌4x+cos4x |
‌dx ‌=−‌‌‌| sin‌x‌cos‌x |
| sin‌4x+cos4x |
‌dx‌=−‌‌‌| sin‌x‌cos‌x‌dx |
| 1−2sin‌2x×cos2x |
‌=−‌‌‌‌dx‌=−‌‌‌‌dxLet
cos‌2‌x=t
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