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प्रश्न
Evaluate the following:
`int_0^(pi/6) sin^5 3x "d"x`
उत्तर
Put t = 3x
dt = 3 dx
`"dt"/3` = dx
I = `int_0^(pi/2) sin^5"t" "dt"/3`
| x | 0 | `pi/6` |
| t | 0 | `pi/2` |
Here n = 5, which is odd
∴ `int_^(pi/2) sin^"n" x "d"x = (("n" - 1))/"n" xx (("n" - 3))/(("n" - 2)) xx ..... xx 2/3`
`int_0^(pi/2) sin^5 "t" "dt"/3 = 1/3 xx 4/5 xx 2/3 xx 1`
I = `8/45`
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