# Thread: integration of area of inverse trig

1. ## integration of area of inverse trig

find the area between the curve y=tan-1x, the X axis and the line x=sqrt3

find the area of the region enclosed by the curves
y=2sin[(pi)x/4] and x=2sin[(pi)y/4]

thankyou =]

2. Originally Posted by christina
find the area between the curve y=tan-1x, the X axis and the line x=sqrt3
This is, of course, $\int_{x=0}^{\sqrt{3}} tan^{-1}(x) dx$. To do that integral, use integration by parts with $u= tan^{-1}(x)$, dv= dx.

find the area of the region enclosed by the curves
y=2sin[(pi)x/4] and x=2sin[(pi)y/4]
These two curves intersect at (0,0) and (2,2) so this is [tex]\int_{x=0}^2 (2sin(\pi x/4)- 4 sin^{-1}(x/2)/4) dx.
You should be able to look up the integrals of both those functions.

thankyou =]

3. Originally Posted by christina
find the area between the curve y=tan-1x, the X axis and the line x=sqrt3

find the area of the region enclosed by the curves
y=2sin[(pi)x/4] and x=2sin[(pi)y/4]

thankyou =]
For #1

Graphical hint