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
Solve 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
The value of \[\int\limits_0^{\pi/2} \cos x\ e^{\sin x}\ dx\] is
Evaluate : \[\int\limits_0^\pi/4 \frac{\sin x + \cos x}{16 + 9 \sin 2x}dx\] .
\[\int\limits_1^5 \frac{x}{\sqrt{2x - 1}} dx\]
\[\int\limits_0^1 \tan^{- 1} x dx\]
\[\int\limits_0^1 \log\left( 1 + x \right) dx\]
Evaluate the following integrals :-
\[\int_2^4 \frac{x^2 + x}{\sqrt{2x + 1}}dx\]
\[\int\limits_0^1 \left( \cos^{- 1} x \right)^2 dx\]
\[\int\limits_0^{\pi/4} \tan^4 x dx\]
\[\int\limits_0^{\pi/2} \frac{1}{1 + \tan^3 x} dx\]
\[\int\limits_0^{\pi/2} \frac{x}{\sin^2 x + \cos^2 x} dx\]
\[\int\limits_0^\pi \frac{dx}{6 - \cos x}dx\]
\[\int\limits_0^2 \left( x^2 + 2 \right) dx\]
\[\int\limits_0^3 \left( x^2 + 1 \right) dx\]
Using second fundamental theorem, evaluate the following:
`int_1^2 (x "d"x)/(x^2 + 1)`
Evaluate the following integrals as the limit of the sum:
`int_1^3 x "d"x`
`int x^9/(4x^2 + 1)^6 "d"x` is equal to ______.
