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Question
By using the properties of the definite integral, evaluate the integral:
`int_0^(2x) cos^5 xdx`
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Solution
Let f (x) = cos5 x
Now we have
f (2π - x) = (cos (2π - x))5
= (cos x)5 = cos5 x = f (x)
⇒ `I = 2 int_0^pi cos^5 x dx`
`[∵ int_0^(2a) f (x) dx = 2 int_0^a f (x)dx, if (2a - x) = f(x) = 0, if (2a - x) = -f(x)]`
Again, we have
f (π - x) = (cos (π - x))5 = -cos5 x = - f(x)
⇒ `2 int_0^pi cos^5 x dx = 0`
Hence, `int_0^(2pi) cos^5 x dx `
`= 2 int_0^5 cos^5 x dx `
= 2 × 0
= 0
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