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
\[\int\limits_{\pi/4}^{\pi/2} \cot x\ dx\]
Evaluate the following integral:
Evaluate each of the following integral:
\[\int_a^b \frac{x^\frac{1}{n}}{x^\frac{1}{n} + \left( a + b - x \right)^\frac{1}{n}}dx, n \in N, n \geq 2\]
If f(2a − x) = −f(x), prove that
Evaluate each of the following integral:
The value of \[\int\limits_0^{2\pi} \sqrt{1 + \sin\frac{x}{2}}dx\] is
If \[I_{10} = \int\limits_0^{\pi/2} x^{10} \sin x\ dx,\] then the value of I10 + 90I8 is
\[\int\limits_0^1 \cos^{- 1} \left( \frac{1 - x^2}{1 + x^2} \right) dx\]
\[\int\limits_0^1 \frac{1 - x}{1 + x} dx\]
\[\int\limits_0^\pi \sin^3 x\left( 1 + 2 \cos x \right) \left( 1 + \cos x \right)^2 dx\]
\[\int\limits_0^\infty \frac{x}{\left( 1 + x \right)\left( 1 + x^2 \right)} dx\]
\[\int\limits_0^{\pi/2} \frac{x}{\sin^2 x + \cos^2 x} dx\]
Using second fundamental theorem, evaluate the following:
`int_0^3 ("e"^x "d"x)/(1 + "e"^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 (2x + 3) "d"x`
Choose the correct alternative:
`int_(-1)^1 x^3 "e"^(x^4) "d"x` is
Find `int x^2/(x^4 + 3x^2 + 2) "d"x`
Which notation represents a definite integral?
Which integral has lower and upper limits?
What is an antiderivative of \[2x+1\]?
