Advertisements
Advertisements
प्रश्न
Advertisements
उत्तर
\[\int_0^3 \frac{1}{x^2 + 9} d x\]
\[ = \int_0^3 \frac{1}{x^2 + 3^2} d x\]
\[ = \frac{1}{3} \left[ \tan^{- 1} \frac{x}{3} \right]_0^3 \]
\[ = \frac{1}{3}\left( \tan^{- 1} 1 - \tan^{- 1} 0 \right)\]
\[ = \frac{1}{3}\left( \frac{\pi}{4} - 0 \right)\]
\[ = \frac{\pi}{12}\]
APPEARS IN
संबंधित प्रश्न
If f is an integrable function, show that
\[\int\limits_0^\infty \frac{1}{1 + e^x} dx\] equals
The value of \[\int\limits_0^{\pi/2} \cos x\ e^{\sin x}\ dx\] is
If \[\int\limits_0^1 f\left( x \right) dx = 1, \int\limits_0^1 xf\left( x \right) dx = a, \int\limits_0^1 x^2 f\left( x \right) dx = a^2 , then \int\limits_0^1 \left( a - x \right)^2 f\left( x \right) dx\] equals
\[\int\limits_0^1 \left( \cos^{- 1} x \right)^2 dx\]
\[\int\limits_0^{\pi/4} e^x \sin x dx\]
\[\int\limits_{- a}^a \frac{x e^{x^2}}{1 + x^2} dx\]
\[\int\limits_0^{\pi/2} \frac{1}{1 + \cot^7 x} dx\]
\[\int\limits_0^\pi \frac{x \sin x}{1 + \cos^2 x} dx\]
\[\int\limits_0^\pi \frac{x}{a^2 - \cos^2 x} dx, a > 1\]
\[\int\limits_{- \pi}^\pi x^{10} \sin^7 x dx\]
\[\int\limits_0^1 \cot^{- 1} \left( 1 - x + x^2 \right) dx\]
\[\int\limits_0^2 \left( 2 x^2 + 3 \right) dx\]
Find : `∫_a^b logx/x` dx
Using second fundamental theorem, evaluate the following:
`int_0^1 "e"^(2x) "d"x`
Verify the following:
`int (x - 1)/(2x + 3) "d"x = x - log |(2x + 3)^2| + "C"`
`int x^3/(x + 1)` is equal to ______.
