Question

Evaluate the following integrals using properties of integration:

`int_0^pi x[sin^2(sin x) cos^2 (cos x)] "d"x`

Answer 1

Let I = `int_0^pi x[sin^2(sin x) cos^2 (cos x)] "d"x`

fx) = sin²(sin x) + cos²(cos x)

f(π – x) = sin²(sin π – x)) + cos²(cos(π – x))

= sin²(sin x) + cos²(cos x)

f(x) = f(π – x)

`int_0^"a" x f(x)  "d"x = "a"/ int_0^"a" f(x  "d"x`

If `"f"("a" - x) = "f"(x)`

∴ I = `pi/2 int_0^pi [sin^2(sin x) + cos^2(cos x)]  "d"x`

If `"f"(2"a" - x) = "f"(x)`, then `int_0^(2"a") f(x)  "d"x = 2int_0^"a" f(x)  "d"x`

I = `pi/2 xx 2 int_0^(pi/2) [sin^2(sin x) + cos^2 (cos x)]  "d"x`  ........(1)

I = `pi int_0^(pi/2) [sin^2(sin(pi/2 - x) + cos^2(cos(pi/2 - ))]  "d"x`

∵ `int_0^"a" f(x)  "d"x = int_0^"a" f("a" - x)  "d"x`

= `pi  int_0^(pi/2) [sin^2 (cos x ) + cos^2 (sin x)]  "d"x`  ........(2)

Add (1) + (2)

2I = `pi  int_0^(pi/2) [sin^2 (sin x) + cos^2 (cos x) + sin^2(cos x) + cos^2 (sin x)]  "d"x`

= `pi int_0^(pi/2) 2  "d"x`

= `pi [2x]_0^(pi/2)`

= `2pi xx pi/2`

= `pi^2`

2I = `pi^2`

I = `pi^2/2`