1use the property
f(a)=f(a+b-x)
u will get ur answer
∫ln(1+tanx) from o to pi/4
1)pi*ln2/8
2)pi*ln2/4
3)pi*ln2
4)pi*ln4
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UP 0 DOWN 0 0 10
10 Answers
1one more pls.............
∫[x2-x+1] where [] denotes integer function from 0 to 2
13pi*ln2/8
1+ tan(pi/4-x)= 2/(1+tanx)
I=0∫pi/4ln2dx - I
2I=pi/4*ln2
1HINT for #3
look when the GINT beomes 1
and when it becomes 2
u have to substitute the values according to that
WARNING : DON'T BLINDLY PUT LIMITS AS 0 TO 1 AND 1TO2
1ya abhi u r rite pls post me the soln
thanks
i got
wehere i commited mistake
11)
u will get
∫0pi/4ln(2/1+tanx) =I
∫0pi/4ln(1+tanx) =I
adding both u get
2I=∫0pi/4 ln2
I=pi*ln2/8
102∫[x2-x+1]
x2-x+1=1
when x=0 and at x=1
x2-x+1=2
x=1/2+√5/2
x2-x+1=3
x=1/2+3/2=2
∫010dx +1/2+√5/21∫2dx +1/2+√5/22∫3dx
u get 2(1/2+√5/2-1)+3(2-1/2+√5/2)
evaluate it to get ur answer