Advertisements
Advertisements
प्रश्न
Advertisements
उत्तर
\[Let I = \int_0^\frac{1}{2} \frac{1}{\sqrt{1 - x^2}} d x . Then, \]
\[I = \left[ \sin^{- 1} x \right]_0^\frac{1}{2} \]
\[ \Rightarrow I = \sin^{- 1} \frac{1}{2} - \sin^{- 1} 0\]
\[ \Rightarrow I = \frac{\pi}{6} - 0\]
\[ \Rightarrow I = \frac{\pi}{6}\]
APPEARS IN
संबंधित प्रश्न
The value of \[\int\limits_0^{\pi/2} \cos x\ e^{\sin x}\ dx\] is
The derivative of \[f\left( x \right) = \int\limits_{x^2}^{x^3} \frac{1}{\log_e t} dt, \left( x > 0 \right),\] is
The value of \[\int\limits_0^{\pi/2} \log\left( \frac{4 + 3 \sin x}{4 + 3 \cos x} \right) dx\] is
\[\int\limits_1^2 x\sqrt{3x - 2} dx\]
\[\int\limits_0^{\pi/4} \sin 2x \sin 3x dx\]
\[\int\limits_0^{\pi/4} \cos^4 x \sin^3 x dx\]
\[\int\limits_0^\pi \frac{x \sin x}{1 + \cos^2 x} dx\]
\[\int\limits_0^\pi x \sin x \cos^4 x dx\]
\[\int\limits_0^\pi \frac{dx}{6 - \cos x}dx\]
\[\int\limits_0^{\pi/2} \frac{1}{2 \cos x + 4 \sin x} dx\]
\[\int\limits_0^{\pi/2} \frac{dx}{4 \cos x + 2 \sin x}dx\]
Using second fundamental theorem, evaluate the following:
`int_0^1 x"e"^(x^2) "d"x`
Evaluate the following:
f(x) = `{{:("c"x",", 0 < x < 1),(0",", "otherwise"):}` Find 'c" if `int_0^1 "f"(x) "d"x` = 2
Integrate `((2"a")/sqrt(x) - "b"/x^2 + 3"c"root(3)(x^2))` w.r.t. x
