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Question : 31 of 150
Marks:
+1,
-0
Solution:
Let
I=∫0log(1+tan‌x)‌d‌x...(i)
⇒I=∫0log[1+tan(−x)]‌d‌x [∵∫0af(x)dx=∫0af(a−x)dx] =∫0log[1+]‌d‌x =∫0log[]‌d‌x =∫0log‌2‌d‌x−∫0log(1+tan‌x)‌d‌x ⇒I=log‌2[x]0−I‌‌‌‌‌‌‌‌‌‌‌ [from Eq. (i)]
⇒2I=loge‌2 ⇒I=loge‌2
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