Integration problem! So confused, inverse trig functions :/

So how on earth would i go about integrating these problems....
1. Integrate from 0 to 1/sqrt2 -----> arcsinx/sqrt(1-x^2)
2. "" "" arccosx/sqrt(1-x^2)

I have no idea where to even start! Mind=BLOWN

The key here is that you should remember something about inverse trig functions from differential calculus: namely their derivatives (which bizarrely turn out to be algebraic functions)

d arcsin(x) / dx = 1/sqrt(1-x^2) and d arccos(x) / dx = -1/sqrt(1-x^2)

Now, regardless of how ugly the functions are, when you see an integrand which is the product of a function and its derivative (and maybe a constant), what technique of integration should you use?

Drum roll please. . . . Substitution! Let the function be u, and its derivative times the dx (or whatever the differential of the original variable was be du), for definite integrals, rewrite the bounds in terms of u, so you don't have to change variable back.

So in your first problem, use u = arcsin(x) and du = dx/sqrt(1-x^2).

By the way, when writing integrals, include the differential dx (or d whatever variable it is): it's not just a reminder of the variable of integration. There is a viewpoint that says one doesn't integrate functions at all, one integrates differentials, and things like substitution and integration by parts are made harder by trying to pretend it's just telling you the variable of integration.

1.


u=sin^(-1) (x)

du = dx/ sqrt(1-x^2)

Change limits

x=0 --> u=0

x=1/ sqrt(2) --> u= pi/ 4

Therefore

= int (0..pi/4) u du

= (1/2) [ (pi/4)]^2 = pi^2 / 32



Similarly for the second one

u=cos^(-1) (x)

du = -dx / sqrt(1-x^2)

Change limits

x=0 --> u=1

x=1/sqrt(2) --> u= pi/4

Therefore

= - int (1..pi/4) u du

= -(1/2) [ (pi/4)^2 -1^2] = -(1/2) [ (pi^2 -16)/16] = (-1/32) [ pi^2 -16]