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
\[\int\limits_{\pi/4}^{\pi/2} \cot x\ dx\]
\[\int\limits_0^1 \left\{ x \right\} dx,\] where {x} denotes the fractional part of x.
The value of \[\int\limits_0^{2\pi} \sqrt{1 + \sin\frac{x}{2}}dx\] is
\[\int\limits_0^\pi \sin^3 x\left( 1 + 2 \cos x \right) \left( 1 + \cos x \right)^2 dx\]
\[\int\limits_0^{\pi/2} \frac{1}{1 + \cot^7 x} dx\]
\[\int\limits_0^{\pi/2} \frac{x \sin x \cos x}{\sin^4 x + \cos^4 x} dx\]
\[\int\limits_0^\pi \frac{x \tan x}{\sec x + \tan x} dx\]
\[\int\limits_2^3 \frac{\sqrt{x}}{\sqrt{5 - x} + \sqrt{x}} dx\]
\[\int\limits_0^{\pi/2} \frac{\sin^2 x}{\sin x + \cos x} dx\]
\[\int\limits_0^{\pi/2} \frac{x}{\sin^2 x + \cos^2 x} dx\]
\[\int\limits_0^2 \left( 2 x^2 + 3 \right) dx\]
Using second fundamental theorem, evaluate the following:
`int_1^2 (x - 1)/x^2 "d"x`
Evaluate the following integrals as the limit of the sum:
`int_0^1 (x + 4) "d"x`
Evaluate the following integrals as the limit of the sum:
`int_1^3 (2x + 3) "d"x`
Evaluate `int "dx"/sqrt((x - alpha)(beta - x)), beta > alpha`
Given `int "e"^"x" (("x" - 1)/("x"^2)) "dx" = "e"^"x" "f"("x") + "c"`. Then f(x) satisfying the equation is:
