Advertisements
Advertisements
प्रश्न
\[\int_0^\frac{\pi^2}{4} \frac{\sin\sqrt{x}}{\sqrt{x}} dx\] equals
विकल्प
2
1
π/4
π2/8
उत्तर
2
\[\int_0^\frac{\pi^2}{4} \frac{\sin\sqrt{x}}{\sqrt{x}} d x\]
\[Let\ \sqrt{x} = t, then\ \frac{1}{2\sqrt{x}}dx = dt\]
\[When\ \ x = 0, t = 0, x = \frac{\pi^2}{4}, t = \frac{\pi}{2}\]
\[\text{Therefore the integral becomes}\]
\[ \int_0^\frac{\pi}{2} 2\ sint\ dt\]
\[ = - 2 \left[ cost \right]_0^\frac{\pi}{2} \]
\[ = 2\]
APPEARS IN
संबंधित प्रश्न
\[\int\limits_1^4 f\left( x \right) dx, where f\left( x \right) = \begin{cases}7x + 3 & , & \text{if }1 \leq x \leq 3 \\ 8x & , & \text{if }3 \leq x \leq 4\end{cases}\]
The value of \[\int\limits_{- \pi/2}^{\pi/2} \left( x^3 + x \cos x + \tan^5 x + 1 \right) dx, \] is
\[\int\limits_0^{\pi/2} \frac{\sin x}{\sqrt{1 + \cos x}} dx\]
\[\int\limits_0^\infty \frac{x}{\left( 1 + x \right)\left( 1 + x^2 \right)} dx\]
\[\int\limits_0^{\pi/4} \sin 2x \sin 3x dx\]
\[\int\limits_{- \pi/2}^{\pi/2} \sin^9 x dx\]
\[\int\limits_0^{15} \left[ x^2 \right] dx\]
\[\int\limits_{- \pi}^\pi x^{10} \sin^7 x dx\]
\[\int\limits_{\pi/6}^{\pi/2} \frac{\ cosec x \cot x}{1 + {cosec}^2 x} dx\]
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")`
Evaluate the following using properties of definite integral:
`int_0^(i/2) (sin^7x)/(sin^7x + cos^7x) "d"x`
Evaluate the following integrals as the limit of the sum:
`int_1^3 x "d"x`
Evaluate the following integrals as the limit of the sum:
`int_1^3 (2x + 3) "d"x`
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
Choose the correct alternative:
If n > 0, then Γ(n) is
Find `int x^2/(x^4 + 3x^2 + 2) "d"x`
If `int (3"e"^x - 5"e"^-x)/(4"e"6x + 5"e"^-x)"d"x` = ax + b log |4ex + 5e –x| + C, then ______.
`int (cos2x - cos 2theta)/(cosx - costheta) "d"x` is equal to ______.
`int x^9/(4x^2 + 1)^6 "d"x` is equal to ______.
`int x^3/(x + 1)` is equal to ______.