Advertisements
Advertisements
Question
Evaluate the following:
`int_0^(pi/2) cos^7 x "d"x`
Solution
Here n = 7, which is odd
∴ `int_0^(pi/2) cos^"n" x "d"x = (("" - 1))/"n" xx (("n" - 3))/(("n" - 2)) xx ...... xx 2/3`
`int_0^(pi/2) cos^7 x "d"x = 6/7 xx 4/5 xx 2/3 xx 1`
= `16/35`
APPEARS IN
RELATED QUESTIONS
Evaluate the following:
`int_0^(pi/2) sin^10 x "d"x`
Evaluate the following:
`int_0^(pi/4) sin^6 2x "d"x`
Evaluate the following:
`int_0^(pi/6) sin^5 3x "d"x`
Evaluate the following:
`int_0^(pi/2) sin^2x cos^4 x "d"x`
Evaluate the following:
`int_0^(2pi) sin^7 x/4 "d"x`
Evaluate the following:
`int_0^(pi/2) sin^3theta cos^5theta "d"theta`
Evaluate the following:
`int_1^0 x^2 (1 - x)^3 "d"x`
Choose the correct alternative:
The value of `int_0^(pi/6) cos^3 3x "d"x` is
Choose the correct alternative:
If `f(x) = int_1^x "e"^(sin u)/u "d"u, x > 1` and `int_1^3 "e"^(sin x^2)/x "d"x = 1/2 [f("a") - f(1)]`. then one of the possible value of a is
Choose the correct alternative:
The value of `int_0^1 (sin^-1x)^2 "d"x` is
Choose the correct alternative:
The value of `int_0^"a" (sqrt("a"^2 - x^2))^3 "d"x` is
