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` is an integrable function such that f(2a − x) = f(x), then prove that
If \[\left[ \cdot \right] and \left\{ \cdot \right\}\] denote respectively the greatest integer and fractional part functions respectively, evaluate the following integrals:
`int_0^1 sqrt((1 - "x")/(1 + "x")) "dx"`
The value of \[\int\limits_{- \pi}^\pi \sin^3 x \cos^2 x\ dx\] is
\[\int\limits_1^5 \frac{x}{\sqrt{2x - 1}} dx\]
\[\int\limits_0^1 \tan^{- 1} \left( \frac{2x}{1 - x^2} \right) dx\]
\[\int\limits_0^{\pi/2} \frac{\sin^2 x}{\left( 1 + \cos x \right)^2} dx\]
\[\int\limits_0^\infty \frac{x}{\left( 1 + x \right)\left( 1 + x^2 \right)} dx\]
\[\int\limits_0^{\pi/2} \frac{\sin^2 x}{\sin x + \cos x} dx\]
\[\int\limits_0^{\pi/2} \frac{dx}{4 \cos x + 2 \sin x}dx\]
Find : `∫_a^b logx/x` dx
Using second fundamental theorem, evaluate the following:
`int_1^2 (x "d"x)/(x^2 + 1)`
Evaluate the following:
`int_1^4` f(x) dx where f(x) = `{{:(4x + 3",", 1 ≤ x ≤ 2),(3x + 5",", 2 < x ≤ 4):}`
Evaluate the following using properties of definite integral:
`int_(-1)^1 log ((2 - x)/(2 + x)) "d"x`
If f(x) = `{{:(x^2"e"^(-2x)",", x ≥ 0),(0",", "otherwise"):}`, then evaluate `int_0^oo "f"(x) "d"x`
Evaluate the following integrals as the limit of the sum:
`int_1^3 x "d"x`
Evaluate `int "dx"/sqrt((x - alpha)(beta - x)), beta > alpha`
Find `int sqrt(10 - 4x + 4x^2) "d"x`
`int (cos2x - cos 2theta)/(cosx - costheta) "d"x` is equal to ______.
