Advertisements
Advertisements
प्रश्न
Advertisements
उत्तर
\[\int_0^{2\pi} \cos^{- 1} \left( \cos x \right)dx\]
\[ = \int_0^\pi \cos^{- 1} \left( \cos x \right)dx +\int_\pi^{2\pi} \cos^{- 1} \left( \cos x \right)dx\]
\[ = \int_0^\pi xdx + \int_\pi^{2\pi} \left( 2\pi - x \right)dx .....................\left[ \pi \leq x \leq 2\pi \Rightarrow - 2\pi \leq - x \leq - \pi \Rightarrow 0 \leq 2\pi - x \leq \pi \right]\]
\[= \left.\frac{x^2}{2}\right|_0^\pi + \left.\frac{\left( 2\pi - x \right)^2}{2 \times \left( - 1 \right)}\right|_\pi^{2\pi} \]
\[ = \frac{1}{2}\left( \pi^2 - 0 \right) - \frac{1}{2}\left( 0 - \pi^2 \right)\]
\[ = \frac{\pi^2}{2} + \frac{\pi^2}{2}\]
\[ = \pi^2\]
APPEARS IN
संबंधित प्रश्न
\[\int\limits_1^4 f\left( x \right) dx, where f\left( x \right) = \begin{cases}7x + 3 & , & \text{if }1 \leq x \leq 3 \\ 8x & , & \text{if }3 \leq x \leq 4\end{cases}\]
Evaluate each of the following integral:
The value of the integral \[\int\limits_0^{\pi/2} \frac{\sqrt{\cos x}}{\sqrt{\cos x} + \sqrt{\sin x}} dx\] is
If \[\int\limits_0^a \frac{1}{1 + 4 x^2} dx = \frac{\pi}{8},\] then a equals
`int_0^(2a)f(x)dx`
\[\int\limits_0^1 x \left( \tan^{- 1} x \right)^2 dx\]
\[\int\limits_0^\pi \frac{x \sin x}{1 + \cos^2 x} dx\]
\[\int\limits_0^\pi x \sin x \cos^4 x dx\]
\[\int\limits_{- \pi}^\pi x^{10} \sin^7 x dx\]
\[\int\limits_0^2 \left( 2 x^2 + 3 \right) dx\]
\[\int\limits_{- 1}^1 e^{2x} dx\]
\[\int\limits_1^3 \left( x^2 + 3x \right) dx\]
Using second fundamental theorem, evaluate the following:
`int_0^1 "e"^(2x) "d"x`
Using second fundamental theorem, evaluate the following:
`int_0^1 x"e"^(x^2) "d"x`
Using second fundamental theorem, evaluate the following:
`int_(-1)^1 (2x + 3)/(x^2 + 3x + 7) "d"x`
Using second fundamental theorem, evaluate the following:
`int_0^(pi/2) sqrt(1 + cos x) "d"x`
Choose the correct alternative:
If n > 0, then Γ(n) is
The value of `int_2^3 x/(x^2 + 1)`dx is ______.
