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Question : 75 of 160
Marks:
+1,
-0
Solution:
Consider the integral.
In=∫dx It is solved as,
In=∫dx =∫dx =∫| sin(n−1)‌x‌cos‌x+cos(n−1)‌x‌sin‌x |
| cos‌x |
dx =∫sin(n−1)‌d‌x+∫| cos(n−1)‌x‌sin‌x |
| cos‌x |
dx Solve further,
I=∫sin(n−1)‌d‌x +‌∫| 2‌sin‌x‌cos(n−1)‌x |
| cos‌x |
dx =−+‌∫| sin‌n‌x+sin(2−n)‌x |
| cos‌x |
dx =−+ ‌∫dx −‌∫dx =−+In−In−2 Solve further,
(1−)In=−−In−2 In=‌cos(n−1)‌x−In−2
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