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Question : 125 of 160
Marks:
+1,
-0
Solution:
Let
I=sin‌(2tan−1√‌)‌dxPut
x=cos‌θ⇒dx=−sin‌θdθ∴‌‌I‌=sin‌(2tan−1√‌)(−sin‌θ)dθ‌=sin‌(2tan−1‌tan(‌−‌))sin‌θdθ‌=sin‌(π−θ)sin‌θdθ=sin‌2θdθ ‌=[‌]dθ=‌[θ−‌]0π∕2‌=‌⌊‌−0⌋=‌
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