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Question : 76 of 160
Marks:
+1,
-0
Solution:
‌‌ Let ‌I=∫−‌‌log(sin‌(4x+3))‌dx...‌ (i) ‌‌‌‌‌‌⇒I=∫−‌‌log(sin‌(4(‌+(−‌)−x)+3))‌dx⇒I=∫−‌‌log(sin‌(‌−(4x+3)))‌dx‌‌⇒I=∫−‌‌log(cos(4x+3))‌dx...‌ (ii) ‌.On adding Eqs. (i) and (ii), we get
2I‌=∫−‌‌log(sin‌(4x+3)‌cos(4x+3))‌dx⇒2I=∫−‌‌log(‌)‌dx⇒2I=∫−‌‌log(sin‌(2(4x+3)))‌dx‌−∫−‌‌(log‌2)‌dx⇒2I=I−(log‌2)[‌−(−‌)]⇒I=−‌‌log‌2
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