integration problem ... ~~ (1 Viewer)

[pc_wizz]

1∫0 [tan-1 (x)] . dx

can anyone help solve this inverse trig integral, it was in a 3U worksheet.

btw the domain of the integral is 1 and 0 ...

thanx

pcwizz

 

[Heinz]

y=arctan(x)
tany=x
@x=0, tany=0 :. y = 0
@x=1, tany=1 :. y= pi/4

Now we find ∫ [tan (y)] . dy between 0 and pi/4. Answer is ln[sqrt(2)]

The area under the inverse tan curve is equal to the total area of the rectangle (1 x pi/4) minus the area under the tany (w.r.t.y).

So we get pi/4 - ln[sqrt(2)] = pi/4 - [ln(2)]/2

4u, just use IBP.

 

[speersy]

whats arc tan(x)?
in 3 unit terms?

 

[Heinz]

Originally posted by speersy
whats arc tan(x)?
in 3 unit terms?

inverse tan. I just used to typing arctan stupid maple...