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By using the properties of the definite integral, evaluate the integral: ∫02xcos5xdx

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प्रश्न

By using the properties of the definite integral, evaluate the integral:

`int_0^(2x) cos^5 xdx`

योग
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उत्तर

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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अध्याय 7: Integrals - Exercise 7.11 [पृष्ठ ३४७]

APPEARS IN

एनसीईआरटी Mathematics Part 1 and 2 [English] Class 12
अध्याय 7 Integrals
Exercise 7.11 | Q 14 | पृष्ठ ३४७

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