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Question : 69 of 160
Marks:
+1,
-0
Solution:
‌‌ Let ‌I=∫‌‌dx‌⇒‌‌I=∫‌‌dx‌⇒‌‌I=∫‌| sin‌5x(1−7cos2x)‌dx |
| sin‌12xcos2x |
Put
sin‌6x‌cos‌x=t‌(6sin‌5xcos2x−sin‌7x)‌dx=dt⇒sin‌5x(6cos2x−sin‌2x)‌dx=dt⇒sin‌5x(6cos2x−1+cos2x)‌dx=dt⇒sin‌5x(1−7cos2x)‌dx=−dt∴‌I=∫‌⇒I=‌+C⇒I=‌+C
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