Advertisements
Advertisements
प्रश्न
Advertisements
उत्तर
\[Let I = \int_\frac{- \pi}{2}^\frac{\pi}{2} \sin^3 x d x\]
\[ = \int_\frac{- \pi}{2}^\frac{\pi}{2} \sin x\left( 1 - \cos^2 x \right)dx\]
\[ = \int_\frac{- \pi}{2}^\frac{\pi}{2} \sin x dx - \int_\frac{- \pi}{2}^\frac{\pi}{2} \sin x \cos^2 x dx\]
\[ = \left[ - \cos x \right]_\frac{- \pi}{2}^\frac{\pi}{2} + \left[ \frac{\cos^3 x}{3} \right]_{- \frac{\pi}{2}}^\frac{\pi}{2} \]
\[ = 0 + 0 = 0\]
APPEARS IN
संबंधित प्रश्न
\[\int\limits_0^1 \left\{ x \right\} dx,\] where {x} denotes the fractional part of x.
The value of the integral \[\int\limits_0^\infty \frac{x}{\left( 1 + x \right)\left( 1 + x^2 \right)} dx\]
\[\int\limits_0^{\pi/2} \frac{\cos x}{1 + \sin^2 x} dx\]
\[\int\limits_1^2 \frac{x + 3}{x\left( x + 2 \right)} dx\]
\[\int\limits_0^a \frac{\sqrt{x}}{\sqrt{x} + \sqrt{a - x}} dx\]
\[\int\limits_0^{\pi/2} \frac{x \sin x \cos x}{\sin^4 x + \cos^4 x} dx\]
\[\int\limits_0^\pi \cos 2x \log \sin x dx\]
\[\int\limits_0^\pi \frac{dx}{6 - \cos x}dx\]
\[\int\limits_0^{\pi/2} \frac{1}{2 \cos x + 4 \sin x} dx\]
\[\int\limits_0^2 \left( 2 x^2 + 3 \right) dx\]
Using second fundamental theorem, evaluate the following:
`int_0^1 "e"^(2x) "d"x`
If f(x) = `{{:(x^2"e"^(-2x)",", x ≥ 0),(0",", "otherwise"):}`, then evaluate `int_0^oo "f"(x) "d"x`
Choose the correct alternative:
`int_0^oo "e"^(-2x) "d"x` is
Choose the correct alternative:
If f(x) is a continuous function and a < c < b, then `int_"a"^"c" f(x) "d"x + int_"c"^"b" f(x) "d"x` is
Verify the following:
`int (x - 1)/(2x + 3) "d"x = x - log |(2x + 3)^2| + "C"`
`int (cos2x - cos 2theta)/(cosx - costheta) "d"x` is equal to ______.
