Advertisements
Advertisements
प्रश्न
Advertisements
उत्तर
\[Let I = \int_{- \frac{\pi}{2}}^\frac{\pi}{2} \sin^3 x\ d x\]
\[ = \int_{- \frac{\pi}{2}}^\frac{\pi}{2} \sin x \sin^2 x\ dx\]
\[ = \int_{- \frac{\pi}{2}}^\frac{\pi}{2} \sin x\left( 1 - \cos^2 x \right) dx\]
\[Let\ \cos x = t, then - \sin x\ dx = dt, \]
\[When\, x \to - \frac{\pi}{2} ; t \to 0\ and\ x \to \frac{\pi}{2} ; t \to 0\]
\[I = \int_0^0 \left( - 1 + t^2 \right) dt\]
\[\]
\[ = 0\]
APPEARS IN
संबंधित प्रश्न
Evaluate the following integral:
If `f` is an integrable function such that f(2a − x) = f(x), then prove that
If f is an integrable function, show that
\[\int\limits_{- a}^a f\left( x^2 \right) dx = 2 \int\limits_0^a f\left( x^2 \right) dx\]
The value of the integral \[\int\limits_0^{\pi/2} \frac{\sqrt{\cos x}}{\sqrt{\cos x} + \sqrt{\sin x}} dx\] is
If f (a + b − x) = f (x), then \[\int\limits_a^b\] x f (x) dx is equal to
\[\int\limits_0^1 \tan^{- 1} x dx\]
\[\int\limits_0^1 \sqrt{\frac{1 - x}{1 + x}} dx\]
\[\int\limits_0^{\pi/4} \tan^4 x dx\]
\[\int\limits_0^{\pi/2} \frac{1}{1 + \cot^7 x} dx\]
\[\int\limits_0^{2\pi} \cos^7 x dx\]
\[\int\limits_{\pi/6}^{\pi/2} \frac{\ cosec x \cot x}{1 + {cosec}^2 x} dx\]
\[\int\limits_0^2 \left( x^2 + 2 \right) dx\]
Using second fundamental theorem, evaluate the following:
`int_1^2 (x - 1)/x^2 "d"x`
Integrate `((2"a")/sqrt(x) - "b"/x^2 + 3"c"root(3)(x^2))` w.r.t. x
Find `int x^2/(x^4 + 3x^2 + 2) "d"x`
Evaluate `int (x^2 + x)/(x^4 - 9) "d"x`
Verify the following:
`int (2x + 3)/(x^2 + 3x) "d"x = log|x^2 + 3x| + "C"`
