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Question : 73 of 160
Marks:
+1,
-0
Solution:
Here,
In=∫(cosnx+sin‌nx)‌dx In=∫cosn−1x‌cos‌x‌dx+∫sin‌n−1xsin‌x‌dx Using by parts,
⇒In=cosn−1sin‌x+∫(n−1)cosn−2xsin‌2x‌dx−sin‌n−1x‌cos‌x+∫(n−1)sin‌n−2xcos2x‌dx ‌⇒In=cosn−1xsin‌x−sin‌n−1x‌cos‌x+∫(n−1)cosn−2x(1−cos2x)‌dx‌‌‌+∫(n−1)sin‌n−2x(1−sin‌2x)‌dx ⇒In=cosn−1xsin‌x−sin‌n−1x‌cos‌x‌+∫(n−1)cosn−2x‌dx−∫(n−1)cosnx‌dx+‌∫(n−1)sin‌n−2x‌dx‌−∫(n−1)sin‌nx‌dx ⇒‌‌In=cosn−1xsin‌x−sin‌n−1x‌cos‌x‌‌+∫(n−1)cosn−2x‌dx+∫(n−1)sin‌n−2x‌dx‌‌−[(n−1)‌∫(cosnx+sin‌nx)‌dx. ⇒In=cosn−1xsin‌x−sin‌n−1‌x‌cos‌x‌+(i1−1)In−−(n−1)In ‌⇒In(1+n−1)−(n−1)In−2‌‌=cosn−1)‌x‌s‌i‌n‌x−sin‌n−1x‌cos‌x ⇒In=cosn−1xsin‌x−sin‌n−1‌x‌cos‌x‌+(n1−1)In,−(n−1)In ‌⇒In(1+n−1)−(n−1)In−2‌‌=cosn+1xsin‌x−sin‌n−1x‌cos‌x ‌‌‌=‌f(x)‌∴f(x)=cosn−2x−sin‌n−2x.
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