Advertisements
Advertisements
प्रश्न
Advertisements
उत्तर
\[Let\ I = \int_0^\frac{\pi}{2} \cos^3 x\ d\ x . Then, \]
\[I = \int_0^\frac{\pi}{2} \cos^2 x \cos\ x\ d\ x\]
\[ \Rightarrow I = \int_0^\frac{\pi}{2} \left( 1 - \sin^2 x \right) \cos x d x\]
\[Let u = \sin x, du = \cos\ x\ dx\]
\[ \Rightarrow I = \int\left( 1 - u^2 \right) du\]
\[ \Rightarrow I = \left[ u - \frac{u^3}{3} \right]\]
\[ \Rightarrow I = \left[ \sin x - \frac{\sin^3 x}{3} \right]_0^\frac{\pi}{2} \]
\[ \Rightarrow I = 1 - \frac{1}{3} - 0\]
\[ \Rightarrow I = \frac{2}{3}\]
APPEARS IN
संबंधित प्रश्न
If f(x) is a continuous function defined on [−a, a], then prove that
Solve each of the following integral:
If \[\int\limits_0^1 \left( 3 x^2 + 2x + k \right) dx = 0,\] find the value of k.
The value of the integral \[\int\limits_0^{\pi/2} \frac{\sqrt{\cos x}}{\sqrt{\cos x} + \sqrt{\sin x}} dx\] is
\[\int\limits_1^2 x\sqrt{3x - 2} dx\]
\[\int\limits_0^{\pi/3} \frac{\cos x}{3 + 4 \sin x} dx\]
\[\int\limits_{- \pi/2}^{\pi/2} \sin^9 x dx\]
\[\int\limits_0^{\pi/2} \frac{1}{1 + \cot^7 x} dx\]
\[\int\limits_0^\pi \frac{x \sin x}{1 + \cos^2 x} dx\]
\[\int\limits_0^\pi x \sin x \cos^4 x dx\]
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")`
Using second fundamental theorem, evaluate the following:
`int_1^2 (x "d"x)/(x^2 + 1)`
Using second fundamental theorem, evaluate the following:
`int_0^(pi/2) sqrt(1 + cos x) "d"x`
Choose the correct alternative:
`int_(-1)^1 x^3 "e"^(x^4) "d"x` is
Integrate `((2"a")/sqrt(x) - "b"/x^2 + 3"c"root(3)(x^2))` w.r.t. x
Find `int sqrt(10 - 4x + 4x^2) "d"x`
`int x^9/(4x^2 + 1)^6 "d"x` is equal to ______.
