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Question : 101 of 150
Marks:
+1,
-0
Solution:
We have
I=sin‌2‌x‌tan−1(sin‌x)‌d‌x I=2‌sin‌x‌cos‌x‌tan−1(sin‌x)‌d‌x Put‌sin‌x=t⇒cos‌x‌d‌x=dt Where,
x=0,t=0 x=,t=1 ∴I=2ttan−1(t)dt ⇒I=2[[tan−1t.]01 −.dt] ⇒I=2[(tan−11.)−(0) −‌(−)dt] ⇒I=2[−[t−tan−1t]01] ⇒I=2[−(1−−0)] ⇒I=2[−+] ⇒I=2[−] ⇒I=−1
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