Advertisements
Advertisements
प्रश्न
\[\int\limits_0^\pi x \sin x \cos^4 x dx\]
Advertisements
उत्तर
\[Let, I = \int_0^\pi x \sin x \cos^4 x d x ............(1)\]
\[ = \int_0^\pi \left( \pi - x \right) \sin\left( \pi - x \right) \cos^4 \left( \pi - x \right) d x \]
\[ = \int_0^\pi \left( \pi - x \right) \sin x \cos^4 x d x ..............(2)\]
Adding (1) and (2)
\[2I = \int_0^\pi \left[ x \sin x \cos^4 x + \left( \pi - x \right) \sin x \cos^4 x \right] d x \]
\[ = \int_0^\pi \left( x + \pi - x \right) \sin x \cos^4 x d x \]
\[ = \pi \int_0^\pi \sin x \cos^4 x d x \]
\[ = \pi \left[ \frac{- \cos^5 x}{5} \right]_0^\pi \]
\[ = \pi\left[ \frac{1}{5} + \frac{1}{5} \right]\]
\[ = \frac{2\pi}{5}\]
\[Hence, I = \frac{\pi}{5}\]
APPEARS IN
संबंधित प्रश्न
If \[f\left( a + b - x \right) = f\left( x \right)\] , then prove that \[\int_a^b xf\left( x \right)dx = \frac{a + b}{2} \int_a^b f\left( x \right)dx\]
\[\int\limits_0^\infty \frac{1}{1 + e^x} dx\] equals
\[\int_0^\frac{\pi^2}{4} \frac{\sin\sqrt{x}}{\sqrt{x}} dx\] equals
The derivative of \[f\left( x \right) = \int\limits_{x^2}^{x^3} \frac{1}{\log_e t} dt, \left( x > 0 \right),\] is
\[\int\limits_0^1 \tan^{- 1} \left( \frac{2x}{1 - x^2} \right) dx\]
\[\int\limits_1^3 \left| x^2 - 2x \right| dx\]
\[\int\limits_{- a}^a \frac{x e^{x^2}}{1 + x^2} dx\]
\[\int\limits_0^\pi \frac{x \tan x}{\sec x + \tan x} dx\]
\[\int\limits_0^{\pi/2} \frac{dx}{4 \cos x + 2 \sin x}dx\]
\[\int\limits_1^3 \left( 2 x^2 + 5x \right) dx\]
Evaluate the following integrals as the limit of the sum:
`int_0^1 x^2 "d"x`
Choose the correct alternative:
If n > 0, then Γ(n) is
Evaluate `int (3"a"x)/("b"^2 + "c"^2x^2) "d"x`
