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प्रश्न
Evaluate the following:
`int_0^(pi/2) sin^3theta cos^5theta "d"theta`
उत्तर
`int_0^(pi/2) (1 - cos^2theta) cos^5theta sintheta "d"theta`
I = `int_0^(pi/2) (cos^5theta - cos^7theta)sintheta "d"theta`
t = cos θ
dt = – sinθ dθ
| x | 0 | `pi/2` |
| t | 1 | 0 |
I = `int_1^0 ("t"^5- "t"^7)(- "dt")`
= `int_0^1 ("t"^5 - "t"^7)("dt")`
= `["t"^6/6 - "t"^8/8]_0^1`
= `1/6 - 1/8`
= `(8 - 6)/48`
`int_0^(pi/2) sin^3theta cos^5theta "d"theta = 1/24`
Aliter Method:
`int_0^(pi/2) sin^3theta cos^5theta "d"theta`
Here m = 3, which is odd
And n = 5, which is odd
`int_0^(pi/2) sin^"m"x cos^"n" x "d"x = ("n" - 1)/("m" + "n") * ("n" - 3)/("m" + "n" - 2) * ("n" - 5)/("m" + "n" - 4) ... 2/("m" + 3)* 1/("m" + 1)`
`int_0^(pi/2) sin^3theta cos^5theta "d"theta = 4/8 xx 2/6 xx 1/4`
= `1/24`
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