Advertisements
Advertisements
प्रश्न
उत्तर
\[\int_0^1 \frac{1}{1 + x^2} d x\]
\[ = \left[ \tan^{- 1} x \right]_0^1 \]
\[ = \tan^{- 1} 1 - \tan^{- 1} 0\]
\[ = \frac{\pi}{4} - 0\]
\[ = \frac{\pi}{4}\]
APPEARS IN
संबंधित प्रश्न
Write the coefficient a, b, c of which the value of the integral
The value of \[\int\limits_{- \pi}^\pi \sin^3 x \cos^2 x\ dx\] is
If f (a + b − x) = f (x), then \[\int\limits_a^b\] x f (x) dx is equal to
\[\int\limits_0^{\pi/2} \frac{\cos x}{1 + \sin^2 x} dx\]
\[\int\limits_{\pi/3}^{\pi/2} \frac{\sqrt{1 + \cos x}}{\left( 1 - \cos x \right)^{5/2}} dx\]
\[\int\limits_0^1 \left( \cos^{- 1} x \right)^2 dx\]
\[\int\limits_0^{\pi/2} \frac{1}{1 + \tan^3 x} dx\]
\[\int\limits_{- \pi/4}^{\pi/4} \left| \tan x \right| dx\]
\[\int\limits_2^3 \frac{\sqrt{x}}{\sqrt{5 - x} + \sqrt{x}} dx\]
\[\int\limits_{- \pi}^\pi x^{10} \sin^7 x dx\]
\[\int\limits_0^\pi \frac{dx}{6 - \cos x}dx\]
Prove that `int_a^b ƒ ("x") d"x" = int_a^bƒ(a + b - "x") d"x" and "hence evaluate" int_(π/6)^(π/3) (d"x")/(1+sqrt(tan "x")`
Using second fundamental theorem, evaluate the following:
`int_0^1 "e"^(2x) "d"x`
Evaluate `int "dx"/sqrt((x - alpha)(beta - x)), beta > alpha`
Given `int "e"^"x" (("x" - 1)/("x"^2)) "dx" = "e"^"x" "f"("x") + "c"`. Then f(x) satisfying the equation is: