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Question : 73 of 160
Marks:
+1,
-0
Solution:
Given,
∫x3sin‌3x‌dx=f(x)‌cos‌3‌x+g(x)sin‌3x+CLet,
I=∫x3sin‌3x‌dx‌=x3‌∫sin‌3x‌dx−∫(3x2x‌∫sin‌3x‌dx)‌dx‌=x3(‌)−∫(3x2×(‌))‌dx‌=−‌‌cos‌3‌x+∫x2‌cos‌3‌x‌dx‌=−‌‌cos‌3‌x+[x2(‌)−∫(2x×‌)‌dx]‌=−‌‌cos‌3‌x+‌sin‌3x−‌‌∫xsin‌3x‌dx‌=−‌‌cos‌3‌x+‌sin‌3x−‌‌[‌‌cos‌3‌x+‌]+C=‌(−‌+‌)‌cos‌3‌x+(‌−‌)sin‌3x+6∴f(x)=−‌+‌‌ and ‌g(x)=‌−‌Now,
27(f(x)+xg(x))‌=27[(−‌+‌)+x(‌−‌)]‌=27[‌−‌x]=6x−2x=4x
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