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Question : 154 of 180
Marks:
+1,
-0
Solution:
‌Let ‌I‌=‌| xsin‌2nx |
| sin‌2nx+cos2nx |
‌dx . . . (i)
⇒‌‌I‌=‌| (π−x)sin‌2n(π−x) |
| sin‌2n(π−x)+cos2n(π−x) |
‌dx‌=‌| (π−x)sin‌2nx |
| sin‌2nx+cos2nx |
‌dx . . . (ii)
Adding Eqs. (i) and (ii), we get
‌2I=π‌‌| sin‌2nx |
| sin‌2nx+cos2nx |
‌dx . . . (iii)
‌‌ Let ‌I1=‌| sin‌2nx |
| sin‌2nx+cos2nx |
‌dx‌⇒I1=2‌‌| sin‌2nx |
| sin‌2nx+cos2nx |
‌dx . . . (iv)
‌⇒I1=2‌‌| sin‌2n(−x) |
| sin‌2n(‌−x)+cos2n(‌−x) |
‌dx‌=2‌‌‌dx . . . (v)
Adding Eqs. (iv) and (v), we get
‌2I1=2‌1‌dx=2⋅‌=π⇒‌‌I1=‌Substituting the value of
I1 in Eq. (iii), we get
2I=(π)(‌)⇒I=‌
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