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Question : 119 of 150
Marks:
+1,
-0
Solution:
We have,
‌∫sin‌{2tan−1√‌}‌dx=Asin‌−1x‌+Bx√1−x2+C‌‌ Let ‌I=∫sin‌{2tan−1√‌}‌dx‌=−∫sin‌{2tan−1√‌}2sin‌2θdθ‌=−∫sin‌{2tan−1√tan2θ}2sin‌2θdθ‌=−∫sin‌{2tan−1‌tan‌θ}2sin‌2θdθ‌=−∫[sin‌(2θ)]2sin‌2θdθ‌=−∫2sin‌22θdθ‌=−∫(1−cos‌4‌θ)dθ‌=−[θ−‌]+C‌=[−‌cos−1x+‌×2sin‌2θ‌cos‌2‌θ]+C‌=[−‌(‌−sin‌−1x)+‌√1−x2×x]+C‌‌=‌sin‌−1x+‌√1−x2+(C−‌)But
‌‌I=Asin‌−1x+Bx√1−x2+C∴‌‌A=‌‌‌ and
‌‌B=‌Hence,
‌‌A+B=‌+‌=1
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