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Question
Evaluate the following integrals using properties of integration:
`int_0^pi x[sin^2(sin x) cos^2 (cos x)] "d"x`
Solution
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`
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