Advertisements
Advertisements
प्रश्न
\[\int\limits_{\pi/3}^{\pi/2} \frac{\sqrt{1 + \cos x}}{\left( 1 - \cos x \right)^{5/2}} dx\]
Advertisements
उत्तर
\[\int_\frac{\pi}{3}^\frac{\pi}{2} \frac{\sqrt{1 + \cos x}}{\left( 1 - \cos x \right)^\frac{5}{2}} d x\]
\[ = \int_\frac{\pi}{3}^\frac{\pi}{2} \frac{\sqrt{1 + \cos x}}{\left( 1 - \cos x \right)^\frac{5}{2}} \times \frac{\sqrt{1 - \cos x}}{\sqrt{1 - \cos x}} d x\]
\[ = \int_\frac{\pi}{3}^\frac{\pi}{2} \frac{\sin x}{\left( 1 - \cos x \right)^3}dx\]
\[ = - \frac{1}{2} \left[ \left( 1 - \cos x \right)^{- 2} \right]_\frac{\pi}{3}^\frac{\pi}{2} \]
\[ = - \frac{1}{2}\left[ 1 - 4 \right]\]
\[ = \frac{3}{2}\]
APPEARS IN
संबंधित प्रश्न
Evaluate the following integral:
If `f` is an integrable function such that f(2a − x) = f(x), then prove that
\[\int\limits_0^\pi \frac{1}{1 + \sin x} dx\] equals
`int_0^1 sqrt((1 - "x")/(1 + "x")) "dx"`
\[\int\limits_0^1 \tan^{- 1} x dx\]
\[\int\limits_1^3 \left| x^2 - 4 \right| dx\]
\[\int\limits_{- 1/2}^{1/2} \cos x \log\left( \frac{1 + x}{1 - x} \right) dx\]
\[\int\limits_0^\pi \frac{x}{1 + \cos \alpha \sin x} dx\]
\[\int\limits_0^\pi \cos 2x \log \sin x dx\]
\[\int\limits_0^\pi \frac{x \tan x}{\sec x + \tan x} dx\]
Find : `∫_a^b logx/x` dx
Using second fundamental theorem, evaluate the following:
`int_0^(1/4) sqrt(1 - 4) "d"x`
Using second fundamental theorem, evaluate the following:
`int_0^3 ("e"^x "d"x)/(1 + "e"^x)`
Using second fundamental theorem, evaluate the following:
`int_1^2 (x - 1)/x^2 "d"x`
Evaluate the following:
`int_0^oo "e"^(-4x) x^4 "d"x`
Evaluate the following integrals as the limit of the sum:
`int_0^1 x^2 "d"x`
Choose the correct alternative:
`Γ(3/2)`
Choose the correct alternative:
`int_0^oo x^4"e"^-x "d"x` is
`int (x + 3)/(x + 4)^2 "e"^x "d"x` = ______.
