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Question : 100 of 100
Marks:
+1,
-0
Solution:
‌‌dx =‌+2‌‌‌dx =0+4‌‌ [ as
f(x)‌dx=0 ]
if
f(x) is odd
=2‌f(x)‌dx if
f(x) is even.
I=4‌‌| (π−x)sin‌(π−x) |
| 1+cos2(π−x) |
‌dx I=4‌‌ ⇒I=4π‌‌−4‌∫‌ ⇒2I=4π‌‌‌dx put
cos‌x=t⇒−sin‌x‌dx=dt ∴I=−2π‌‌‌dt =2π‌‌‌dt =2π[tan−1t]−11 =2π[tan−11−tan−1(−1)] =2π[‌−(‌)]=2π‌=π2
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