IIT JEE , AIEEE Forum - Mathematics , Integral Calculus

aman531

kindly solve this

integration of [1/{1+(tan x)^1/2}]dx limits 0 to pi/2

KAB

_{^{Given integral is I=}[ 0]}^{[pi/2 ]} dx/(1+

tanx)

_^I=[0]

^{[pi/2 ]} dx/(1+

tanx)=_{[ 0]}^{[pi/2 ]}

cosxdx/(

sinx+

cosx).......(1)

Also I=_{[ 0]}^{[pi/2 ]}

cos(pi/2-x)dx/(

sin(pi/2-x)+

cos(pi/2-x))=_{[ 0]}^{[pi/2 ]}

sinxdx/(

sinx+

cosx)........(2) (Using _{[ 0]}^[a]f(x)dx=_{[ 0]}^[a]f(a-x)dx)

Add (1) and (2)

we get I=pi/4

jaskaran

is the answer pi/4

explanation:

write (tanx) as sinx/cosx
then use the property integ f(x) from 0 to a = integ f(a-x) from 0 to a
then add the two.

dpgol88

Ans is pi/4

Let I=_{[0 ]}

^{[pi/2 ]} dx/{1+(tanx)^1/2} ....(i)

=_{[0 ]}

^{[pi/2 ]} dx/{1+(cotx)^1/2} .........(ii) (property of DI)

(i) +(ii)

2I=_{[0 ]}

^{[pi/2 ]} 1 dx (u can add the above 2 steps, i hope)

=pi/2

Hence, I= pi/4