Advertisements
Advertisements
प्रश्न
Advertisements
उत्तर
\[Let\ I = \int_0^\frac{\pi}{2} \sin^3 x\ d\ x\ . Then, \]
\[I = \int_0^\frac{\pi}{2} \sin x \sin^2 x\ d\ x\]
\[ \Rightarrow I = \int_0^\frac{\pi}{2} \sin x \left( 1 - \cos^2 x \right) dx\]
\[Let u = \cos x, du = - \sin\ x\ dx\]
\[ \therefore I = \int - \left( 1 - u^2 \right) du\]
\[ \Rightarrow I = \left[ \frac{u^3}{3} - u \right]\]
\[ \Rightarrow I = \left[ \frac{\cos^3 x}{3} - \cos x \right]_0^\frac{\pi}{2} \]
\[ \Rightarrow I = 0 - \frac{1}{3} + 1\]
\[ \Rightarrow I = \frac{2}{3}\]
APPEARS IN
संबंधित प्रश्न
If f(x) is a continuous function defined on [−a, a], then prove that
Evaluate each of the following integral:
Write the coefficient a, b, c of which the value of the integral
\[\int\limits_0^1 \left\{ x \right\} dx,\] where {x} denotes the fractional part of x.
\[\int\limits_0^{\pi/2} \frac{1}{2 + \cos x} dx\] equals
If \[\int\limits_0^a \frac{1}{1 + 4 x^2} dx = \frac{\pi}{8},\] then a equals
The value of \[\int\limits_0^{\pi/2} \log\left( \frac{4 + 3 \sin x}{4 + 3 \cos x} \right) dx\] is
\[\int\limits_0^1 \tan^{- 1} \left( \frac{2x}{1 - x^2} \right) dx\]
\[\int\limits_0^{\pi/2} \frac{\cos x}{1 + \sin^2 x} dx\]
\[\int\limits_0^{\pi/2} x^2 \cos 2x dx\]
\[\int\limits_0^1 \left| \sin 2\pi x \right| dx\]
\[\int\limits_{- \pi/2}^{\pi/2} \sin^9 x dx\]
\[\int\limits_0^{2\pi} \cos^7 x dx\]
\[\int\limits_0^a \frac{\sqrt{x}}{\sqrt{x} + \sqrt{a - x}} dx\]
\[\int\limits_0^{\pi/2} \frac{1}{1 + \tan^3 x} dx\]
\[\int\limits_0^\pi \frac{x \sin x}{1 + \cos^2 x} dx\]
\[\int\limits_0^\pi \frac{x}{a^2 \cos^2 x + b^2 \sin^2 x} dx\]
\[\int\limits_0^{15} \left[ x^2 \right] dx\]
\[\int\limits_0^\pi \frac{x}{a^2 - \cos^2 x} dx, a > 1\]
\[\int\limits_0^{\pi/2} \frac{1}{2 \cos x + 4 \sin x} dx\]
\[\int\limits_0^2 \left( x^2 + 2 \right) dx\]
Using second fundamental theorem, evaluate the following:
`int_1^2 (x - 1)/x^2 "d"x`
Evaluate the following using properties of definite integral:
`int_(- pi/4)^(pi/4) x^3 cos^3 x "d"x`
Evaluate the following:
Γ(4)
Evaluate the following integrals as the limit of the sum:
`int_0^1 (x + 4) "d"x`
