Advertisements
Advertisements
प्रश्न
\[\int\limits_0^\pi \frac{x \sin x}{1 + \cos^2 x} dx\]
Advertisements
उत्तर
\[Let, I = \int_0^\pi \frac{x \sin x}{1 + \cos^2 x} d x ...........(1)\]
\[ = \int_0^\pi \frac{\left( \pi - x \right) \sin\left( \pi - x \right)}{1 + \cos^2 \left( \pi - x \right)} d x\]
\[ = \int_0^\pi \frac{\left( \pi - x \right) \sin x}{1 + \cos^2 x} d x ................(2)\]
Adding (1) and (2)
\[2I = \int_0^\pi \left[ \frac{x \sin x}{1 + \cos^2 x} + \frac{\left( \pi - x \right) \sin x}{1 + \cos^2 x} \right] d x\]
\[ = \int_0^\pi \frac{\pi \sin x}{1 + \cos^2 x} d x \]
\[ = \pi \left[ - \tan^{- 1} \left( cosx \right) \right]_0^\pi \]
\[ = - \pi\left[ \tan^{- 1} \left( - 1 \right) - \tan^{- 1} \left( 1 \right) \right]\]
\[ = - \pi\left( - \frac{\pi}{4} - \frac{\pi}{4} \right)\]
\[ = \frac{\pi^2}{2}\]
\[Hence, I = \frac{\pi^2}{4}\]
APPEARS IN
संबंधित प्रश्न
Evaluate the following integral:
Evaluate the following integral:
If \[\int\limits_0^1 \left( 3 x^2 + 2x + k \right) dx = 0,\] find the value of k.
If \[\int_0^a \frac{1}{4 + x^2}dx = \frac{\pi}{8}\] , find the value of a.
\[\int\limits_1^2 \frac{1}{x^2} e^{- 1/x} dx\]
\[\int\limits_0^{\pi/2} x^2 \cos 2x dx\]
\[\int\limits_0^{\pi/4} \tan^4 x dx\]
\[\int\limits_1^3 \left| x^2 - 4 \right| dx\]
\[\int\limits_0^{\pi/2} \frac{1}{1 + \tan^3 x} dx\]
\[\int\limits_0^1 \cot^{- 1} \left( 1 - x + x^2 \right) dx\]
\[\int\limits_1^3 \left( 2 x^2 + 5x \right) dx\]
\[\int\limits_1^3 \left( x^2 + 3x \right) dx\]
\[\int\limits_0^2 \left( x^2 + 2 \right) dx\]
\[\int\limits_0^3 \left( x^2 + 1 \right) dx\]
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_0^3 ("e"^x "d"x)/(1 + "e"^x)`
Choose the correct alternative:
`int_(-1)^1 x^3 "e"^(x^4) "d"x` is
Choose the correct alternative:
If n > 0, then Γ(n) is
Verify the following:
`int (x - 1)/(2x + 3) "d"x = x - log |(2x + 3)^2| + "C"`
