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Question : 69 of 160
Marks:
+1,
-0
Solution:
We have,
I‌=∫‌| esin‌x(sin‌2x−8‌cos‌x) |
| 2(sin‌x−3)2 |
‌dx⇒‌‌I‌=∫‌| esin‌x(2sin‌x−8)‌cos‌x |
| 2(sin‌x−3)2 |
‌dxPut
sin‌x=t⇒cos‌x‌dx=dt⇒‌I=∫‌‌dt‌I=∫et(‌−‌)‌dt‌[∵∫ex[f(x)+f′(x)]‌dx=exf(x)+C]⇒‌I=et⋅‌+C⇒‌I=esin‌x‌+C
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