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Question : 76 of 160
Marks:
+1,
-0
Solution:
sinmx⋅cos4xdx= ∵sin^nx⋅cosmxdx [(n−1)(n−3)(n−5)...1‌ or ‌2]x =| [(m−1)(m−3)(m−5)...1‌ or ‌2] |
| (m+n)(m+n−2)...1‌ or 2 |
×k Where,
k=‌Both‌m‌and‌n‌are‌even I,otherwise I=sinmxcosnxdx If
m=8,n=4, then
I=[| (7â‹…5â‹…3â‹…1)(3â‹…1) |
| 12â‹…10â‹…8â‹…6â‹…4â‹…2 |
]⋅= ∴m=8
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