Question

Evaluate the integral. ?

Answer 1

0

Explanation:

The derivative of `x^48` in `cos(x^48)` is `48x^47` (power rule), and there is a `x^47` multiplied to the `cos(x^48)` in the integral. This means you can easily use reverse chain rule to integrate.

Rewrite `int_0^(pi^(1/48))x^47*cos(x^48)dx` as `1/48*int_0^(pi^(1/48))48x^47*cos(x^48)dx` and ignore the `1/48` for now.

Since `d/dx(f(g(x))) = f'(g(x))g'(x)` (product rule), `intf'(g(x))g'(x) = f(g(x))`. In this case, `f'(x) = cos(x)` and `g(x) = x^48`.

You can see that `f'(g(x)) = cos(x^48)` and `g'(x) = 48x^47`. Since `f'(x) = cos(x)`, `f(x)` must be `sin(x)` (trig derivatives).

Simply follow the formula above: `1/48*int_0^(pi^(1/48))48x^47*cos(x^48)dx = F(pi^(1/48))-F(0)` where `F(x) = sin(x^48)` (first fundamental theorem of calculus).

Simplify: `F(pi^(1/48))-F(0) = sin((pi^(1/48))^48)-sin(0^48) = sin(pi) - sin(0) = 0 - 0 = 0`