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Question : 31 of 160
Marks:
+1,
-0
Solution:
I=sin−1(‌)dx When,
‌‌x=tan‌θ ‌=‌=sin‌2‌θ Also,
‌‌dx=sec2θdθ and,
‌‌ When
x=0,θ=0 When
‌‌x=1,θ=‌ So,
‌‌I=sin−1(sin‌2‌θ)⋅sec2θ⋅dθ By parts we get,
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