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Question : 76 of 160
Marks:
+1,
-0
Solution:
‌I=‌| log‌2−log(1+x) |
| √1−x2 |
‌dx ‌⇒I=‌‌dx−‌‌dx ‌⇒I=[log‌2‌s‌i‌n‌−1x]−11−∫‌‌dx ‌=π‌log‌2−I1 ‌I1=‌‌dx ‌‌ Put ‌dx=sin‌θ ‌⇒I1=‌‌‌cos‌θ‌d‌θ ‌⇒I1=‌log(1+sin‌θ)‌d‌θ‌‌‌⋅⋅⋅⋅⋅⋅⋅(i) King's property,
⇒‌‌I1‌=π∕2log(1−sin‌θ)‌d‌θ‌‌‌⋅⋅⋅⋅⋅⋅⋅(ii) ⇒‌‌2I1‌=‌log(1−sin‌2θ)‌d‌θ ⇒‌‌I1‌=‌log‌cos‌θ‌d‌θ ‌=2‌log‌cos‌θ‌d‌θ ‌=2(‌‌log‌2) ‌=−π‌log‌2 ∴‌‌I‌=2π‌log‌2
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