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Question : 73 of 160
Marks:
+1,
-0
Solution:
Consider the integral.
I= The above integral is solved as
I=| (cos‌x)√2018 |
| (cos‌x)√2018+(sin‌x)√2018 |
dx…… (I)
Also,
I=| (sin‌x)√2018 |
| (sin‌x)√2018+(cos‌x)√2018 |
dx ……. (II)
Add equation (I) and (II).
2I=| (cos‌x)√2018+(sin‌x)√2018 |
| (cos‌x)√2018+(sin‌x)√2018 |
dx 2I=1dx I= =
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