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Question : 156 of 180
Marks:
+1,
-0
Solution:
‌‌ ‌‌ LHS ‌‌=∫‌| cos‌8‌x+1 |
| tan‌2‌x−cot‌2‌x |
‌dx‌=∫‌| 2cos24x |
(‌| sin‌22x−cos22x | | sin‌2x‌cos‌2‌x |
) |
‌dx‌[∵cos‌2‌x=2cos2x−1]‌=−∫‌| cos24x(2sin‌2x‌cos‌2‌x) |
| (cos22x−sin‌22x) |
‌dx‌=−∫‌| cos24x×sin‌4x |
| cos‌4‌x |
‌dx [∵sin‌2x=2sin‌x‌cos‌x]‌=−‌‌∫2sin‌4x‌cos‌4‌x‌dx‌=−‌‌∫sin‌8x‌dx=‌×‌+c‌‌ Now, ‌‌‌+c=a‌cos‌8‌x+c‌∴‌‌a=‌
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