inverse integral trick!!!!!!

Quite easy...... f (x)=x+sin (x) f'(x)=1+cos (x) we want integral (0 to pi)[f^-1 (x) dx] x=f (t) => dx=f'(t) dt therefore integral becomes Integral( f^-1 (0) to f^-1 (pi)) [ t(1+cos (t)) dt] => Integral (0 to pi) [t+t cos (t) dt] => t^2/2 + t sin (t) + cos (t) | 0 to pi = pi^2/2 - 2 = 2.9348022. .....

It was very easy, as it was me who posted it.....:P

Same range of ever increasing from 0 to Pi for both x and y is the trick.

y = x + Sin x = f (x)

f^-1 (y) = x

d y = (1 + Cos x) d x

Integrate [f^-1 (x) d x, 0 to Pi]

= Integrate [f^-1 (y) d y, 0 to Pi]

= Integrate [x (1 + Cos x) d x, 0 to Pi]

= Pi^2/ 2 + Integrate [x Cos x d x, 0 to Pi]

Let u = x => d u = d x

and d v = Cos x d x => v = Sin x

Pi^2/ 2 + Integrate [x Cos x d x, 0 to Pi]

= Pi^2/ 2 + [x Sin x] of 0 to Pi - Integrate [Sin x d x, 0 to Pi]

= Pi^2/ 2 + [Cos x] of 0 to Pi

= Pi^2/ 2 - [1 – (-1)]

Integrate [f^-1 (x) d x, 0 to Pi] = Pi^2/ 2 - 2 =

2.9348022005446793094172454999381 {The wanted answer}

On the other hand, for y = x + Sin x = f (x)

Integrate [x + Sin x d x, 0 to Pi]

= Pi^2/2 + [-Cos x] of 0 to Pi

= Pi^2/ 2 + [1 – (-1)]

Integrate [f (x) d x, 0 to Pi] = Pi^2/ 2 + 2 =

6.9348022005446793094172454999381

Only y = x + x Cos x is having different range verse domain.

Looking at the graph and by area arguments,

$\displaystyle\int_a^b f(x) dx+\displaystyle\int_{f(a)}^{f(b)} f^{-1} (x) dx=bf(b)-af(a)$

Substitute $a=0,b=\pi,f(x)=x+\sin x$ to get $\displaystyle\int_{0}^{\pi} f^{-1} (x) dx=2.935$

simple approach can be visualizing y axis as x axis and x axis as y axis for integrating any inverse function( if it exists). Then, form a rectangle for the specified interval and subtract the area of that particular function within the interval! this works because inverse for a function is nothing but interchanging the axes !