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Question : 154 of 160
Marks:
+1,
-0
Solution:
‌ Let ‌I‌=‌dθ . . . (i)
‌=‌| (π−θ)sin‌(π−θ) |
| 1+cos2(π−θ) |
dθ⇒I‌=‌| (π−θ)sin‌θ |
| 1+cos2θ |
dθ . . . (ii)
On adding Eqs. (i) and (ii), we get
2I=‌dθLet
cos‌θ=t⇒−sin‌θdθ=dt∴‌‌2I‌=−π‌‌‌dt‌=2π‌‌‌dt‌=2π[tan−1t]01=2π⋅‌ ⇒‌‌I=‌
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