Advertisements
Advertisements
प्रश्न
Advertisements
उत्तर
\[\int_{- \frac{\pi}{2}}^\pi \sin^{- 1} \left( \sin x \right)dx\]
\[ = \int_{- \frac{\pi}{2}}^\frac{\pi}{2} \sin^{- 1} \left( \sin x \right)dx + \int_\frac{\pi}{2}^\pi \sin^{- 1} \left( \sin x \right)dx\]
\[ = \int_{- \frac{\pi}{2}}^\frac{\pi}{2} xdx + \int_\frac{\pi}{2}^\pi \left( \pi - x \right)dx ..............\left[ \frac{\pi}{2} \leq x \leq \pi \Rightarrow - \pi \leq - x \leq - \frac{\pi}{2} \Rightarrow 0 \leq \pi - x \leq \frac{\pi}{2} \right]\]
\[ = \left.\frac{x^2}{2}\right|_{- \frac{\pi}{2}}^\frac{\pi}{2} + \left.\frac{\left( \pi - x \right)^2}{2 \times \left( - 1 \right)}\right|_\frac{\pi}{2}^\pi \]
\[ = \frac{1}{2}\left( \frac{\pi^2}{4} - \frac{\pi^2}{4} \right) - \frac{1}{2}\left( 0 - \frac{\pi^2}{4} \right)\]
\[= 0 + \frac{\pi^2}{8}\]
\[ = \frac{\pi^2}{8}\]
APPEARS IN
संबंधित प्रश्न
If \[f\left( a + b - x \right) = f\left( x \right)\] , then prove that \[\int_a^b xf\left( x \right)dx = \frac{a + b}{2} \int_a^b f\left( x \right)dx\]
If f is an integrable function, show that
\[\int\limits_{- a}^a f\left( x^2 \right) dx = 2 \int\limits_0^a f\left( x^2 \right) dx\]
Evaluate each of the following integral:
Write the coefficient a, b, c of which the value of the integral
\[\int\limits_0^\pi \frac{1}{1 + \sin x} dx\] equals
\[\int_0^\frac{\pi^2}{4} \frac{\sin\sqrt{x}}{\sqrt{x}} dx\] equals
\[\int\limits_0^1 \tan^{- 1} x dx\]
\[\int\limits_0^\pi \sin^3 x\left( 1 + 2 \cos x \right) \left( 1 + \cos x \right)^2 dx\]
\[\int\limits_0^{\pi/4} \tan^4 x dx\]
\[\int\limits_2^3 \frac{\sqrt{x}}{\sqrt{5 - x} + \sqrt{x}} dx\]
\[\int\limits_0^{\pi/2} \frac{1}{2 \cos x + 4 \sin x} dx\]
\[\int\limits_0^4 x dx\]
Using second fundamental theorem, evaluate the following:
`int_0^1 "e"^(2x) "d"x`
Evaluate the following:
`int_1^4` f(x) dx where f(x) = `{{:(4x + 3",", 1 ≤ x ≤ 2),(3x + 5",", 2 < x ≤ 4):}`
Evaluate `int (3"a"x)/("b"^2 + "c"^2x^2) "d"x`
If velocity changes with time, what does a definite integral give over a time interval?
