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Question : 73 of 160
Marks:
+1,
-0
Solution:
Given,
I=∫‌√‌‌dxPut
x=cos2(2θ)dx‌=−sin‌(4θ)×2dθI‌=∫‌×‌×−4sin‌2θ‌cos‌2‌θ‌d‌θ=‌−∫‌| 18sin‌2θ⋅cos‌θ⋅cos‌2‌θ‌d‌θ |
| cos‌θ⋅cos2(2θ) |
dθ=‌+4‌∫‌| cos‌2‌θ−1 |
| cos‌2‌θ |
dθ=‌4θ−4‌∫ sec2θdθ=‌4θ−‌4‌ln| sec2θ+tan‌2‌θ‌+C=‌2‌ln‌|‌|+4θ+C=‌2‌ln‌|‌|‌+2×(‌−sin‌−1√x)+C=‌2‌ln‌|‌|−2sin‌−1√x+CSo,
f(x)= sech−1√x
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