Advertisements
Advertisements
प्रश्न
Advertisements
उत्तर
\[\text{We have}, \]
\[I = \int_{- \frac{\pi}{2}}^\frac{\pi}{2} x \cos^2 x\ d x\]
\[Let f\left( x \right) = x \cos^2 x\]
\[ \Rightarrow f\left( - x \right) = \left( - x \right) \cos^2 \left( - x \right)\]
\[ = - x \cos^2 x\]
\[ \therefore f\left( - x \right) = - f\left( x \right)\]
\[i . e . , f\left( x \right) \text{is odd function}\]
\[\text{We know that} \int_{- a}^a f\left( x \right) d x = 0 , \text{if }f\left( x \right) \text{is odd function} . \]
\[ \therefore I = \int_{- \frac{\pi}{2}}^\frac{\pi}{2} x \cos^2 x\ d x = 0\]
APPEARS IN
संबंधित प्रश्न
Evaluate the following integral:
If f (x) is a continuous function defined on [0, 2a]. Then, prove that
\[\int_0^\frac{\pi^2}{4} \frac{\sin\sqrt{x}}{\sqrt{x}} dx\] equals
The value of the integral \[\int\limits_{- 2}^2 \left| 1 - x^2 \right| dx\] is ________ .
\[\int\limits_0^4 x\sqrt{4 - x} dx\]
\[\int\limits_0^{1/\sqrt{3}} \tan^{- 1} \left( \frac{3x - x^3}{1 - 3 x^2} \right) dx\]
\[\int\limits_0^{\pi/3} \frac{\cos x}{3 + 4 \sin x} dx\]
\[\int\limits_0^{\pi/2} \frac{\sin^2 x}{\left( 1 + \cos x \right)^2} dx\]
\[\int\limits_0^\infty \frac{x}{\left( 1 + x \right)\left( 1 + x^2 \right)} dx\]
\[\int\limits_0^1 \left( \cos^{- 1} x \right)^2 dx\]
\[\int\limits_{- \pi/2}^{\pi/2} \sin^9 x dx\]
\[\int\limits_{- a}^a \frac{x e^{x^2}}{1 + x^2} dx\]
\[\int\limits_0^{\pi/2} \frac{\sin^2 x}{\sin x + \cos x} dx\]
\[\int\limits_2^3 e^{- x} dx\]
Using second fundamental theorem, evaluate the following:
`int_1^"e" ("d"x)/(x(1 + logx)^3`
Evaluate the following using properties of definite integral:
`int_0^(i/2) (sin^7x)/(sin^7x + cos^7x) "d"x`
Evaluate the following:
Γ(4)
Evaluate the following:
`Γ (9/2)`
Choose the correct alternative:
Γ(n) is
If x = `int_0^y "dt"/sqrt(1 + 9"t"^2)` and `("d"^2y)/("d"x^2)` = ay, then a equal to ______.
