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Question : 77 of 160
Marks:
+1,
-0
Solution:
x2sin‌2x‌dx,I=∫x2sin‌2x‌dxPut
t=2x⇒dt=2⋅dxSo,
I=∫‌‌dt=‌‌∫t2sin‌t‌dt‌=‌[t2⋅(−cos‌t)−1⋅(−2)⋅∫t‌cos‌t‌dt]‌=‌[−t2‌cos‌t+2‌∫t⋅cos‌t‌dt]‌=‌[−t2‌cos‌t+2(t⋅sin‌t−(−cos‌t))]‌=‌[−t2‌cos‌t+2tsin‌t+2‌cos‌t]‌=‌[(2x)2‌cos(2x)+2(2x⋅sin‌(2x))+2‌cos(2x)]‌=‌| −x2‌cos(2x)+xsin‌(2x) |
| 2 |
+‌So,
x2sin‌2x‌dx‌=[‌| −x2‌cos(2x)+xsin‌(2x) |
| 2 |
+‌]0‌‌={‌| −()2‌cos(2×)+sin‌(2×) |
| 2 |
+‌}−{‌+‌}‌⇒(0+‌⋅1+‌⋅0)−(0+‌)‌‌‌=‌−‌=‌
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