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Question : 72 of 160
Marks:
+1,
-0
Solution:
‌‌ Let ‌I=∫‌| sin‌2x |
| sin‌2x+3‌cos‌x−3 |
‌dx‌=−∫‌| sin‌2x |
| 1−cos2x+3‌cos‌x−3 |
‌dx‌=∫‌| 2sin‌x⋅cos‌x |
| −cos2x+3‌cos‌x−2 |
‌dx‌=∫‌‌‌‌ Put ‌cos‌x=t‌sin‌x‌dx=−dt‌=∫‌=∫‌‌dt‌=2‌∫(‌−‌)‌dt‌=2[2‌ln(t−2)−ln(t−1)]‌=2‌ln(‌)=ln‌+C‌=ln(‌| (cos‌x−2)4 |
| (cos‌x−1)2 |
)+C
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