Advertisements
Advertisements
प्रश्न
\[\int\limits_0^{2\pi} \cos^7 x dx\]
उत्तर
\[Let, I = \int_0^{2\pi} \cos^7 x d x ..............(1)\]
\[ = \int_0^{2\pi} \cos^7 \left( 2\pi - x \right) d x\]
\[ = \int_0^{2\pi} - \cos^7 x d x\]
\[ \Rightarrow I = - \int_0^{2\pi} \cos^7 x d x ..............(2)\]
Adding (1) and (2) we get,
\[ 2I = \int_0^{2\pi} \cos^7 x d x - \int_0^{2\pi} \cos^7 x d x\]
\[ \Rightarrow 2I = 0\]
\[ \therefore I = 0\]
APPEARS IN
संबंधित प्रश्न
Evaluate each of the following integral:
\[\int\limits_0^\pi \frac{1}{1 + \sin x} dx\] equals
Evaluate : \[\int\limits_0^\pi \frac{x}{1 + \sin \alpha \sin x}dx\] .
\[\int\limits_0^{\pi/2} \frac{\sin^2 x}{\left( 1 + \cos x \right)^2} dx\]
\[\int\limits_1^3 \left| x^2 - 2x \right| dx\]
\[\int\limits_0^\pi \cos 2x \log \sin x dx\]
\[\int\limits_0^\pi \frac{x \tan x}{\sec x + \tan x} dx\]
\[\int\limits_0^1 \cot^{- 1} \left( 1 - x + x^2 \right) dx\]
Using second fundamental theorem, evaluate the following:
`int_1^2 (x "d"x)/(x^2 + 1)`
Using second fundamental theorem, evaluate the following:
`int_0^(pi/2) sqrt(1 + cos x) "d"x`
Evaluate the following:
`int_1^4` f(x) dx where f(x) = `{{:(4x + 3",", 1 ≤ x ≤ 2),(3x + 5",", 2 < x ≤ 4):}`
Evaluate the following:
`int_(-1)^1 "f"(x) "d"x` where f(x) = `{{:(x",", x ≥ 0),(-x",", x < 0):}`
Evaluate the following using properties of definite integral:
`int_(-1)^1 log ((2 - x)/(2 + x)) "d"x`
Evaluate the following integrals as the limit of the sum:
`int_0^1 x^2 "d"x`
Integrate `((2"a")/sqrt(x) - "b"/x^2 + 3"c"root(3)(x^2))` w.r.t. x
`int x^3/(x + 1)` is equal to ______.