© examsiri.com
Question : 155 of 160
Marks:
+1,
-0
Solution:
Let
I=sin‌6xcos5x‌dxI=2‌sin‌6xcos5x‌dx[∵f(2π−x)=f(x)]Let
f(x)=sin‌6xcos5x‌‌‌‌‌‌‌[∵f(π−x)=−f(x)]‌f(π−x)=sin‌6(π−x)⋅cos5(π−x)‌‌=sin‌6x⋅(−cos5x)‌‌=−sin‌6x⋅cos5x‌‌=−f(x)‌∴‌‌I=0
© examsiri.com
Go to Question: