Advertisements
Advertisements
प्रश्न
\[\int\limits_0^{2\pi} \cos^7 x dx\]
Advertisements
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
If f(2a − x) = −f(x), prove that
If \[\int\limits_0^a 3 x^2 dx = 8,\] write the value of a.
Evaluate :
\[\int\limits_1^2 x\sqrt{3x - 2} dx\]
\[\int\limits_0^1 \cos^{- 1} x dx\]
\[\int\limits_0^{\pi/4} e^x \sin x dx\]
\[\int\limits_1^3 \left| x^2 - 2x \right| dx\]
\[\int\limits_0^\pi \frac{x \sin x}{1 + \cos^2 x} dx\]
\[\int\limits_0^{\pi/2} \frac{1}{2 \cos x + 4 \sin x} dx\]
\[\int\limits_0^{\pi/2} \frac{dx}{4 \cos x + 2 \sin x}dx\]
Using second fundamental theorem, evaluate the following:
`int_1^"e" ("d"x)/(x(1 + logx)^3`
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_0^2 "f"(x) "d"x` where f(x) = `{{:(3 - 2x - x^2",", x ≤ 1),(x^2 + 2x - 3",", 1 < x ≤ 2):}`
Evaluate the following using properties of definite integral:
`int_0^1 x/((1 - x)^(3/4)) "d"x`
Evaluate the following:
`int_0^oo "e"^(- x/2) x^5 "d"x`
Evaluate the following integrals as the limit of the sum:
`int_1^3 x "d"x`
Choose the correct alternative:
`int_(-1)^1 x^3 "e"^(x^4) "d"x` is
Choose the correct alternative:
`int_0^oo x^4"e"^-x "d"x` is
Evaluate the following:
`int ((x^2 + 2))/(x + 1) "d"x`
