Advertisements
Advertisements
Question
Solution
\[Let I = \int_0^\frac{1}{2} \frac{1}{\sqrt{1 - x^2}} d x . Then, \]
\[I = \left[ \sin^{- 1} x \right]_0^\frac{1}{2} \]
\[ \Rightarrow I = \sin^{- 1} \frac{1}{2} - \sin^{- 1} 0\]
\[ \Rightarrow I = \frac{\pi}{6} - 0\]
\[ \Rightarrow I = \frac{\pi}{6}\]
APPEARS IN
RELATED QUESTIONS
Evaluate each of the following integral:
If \[\int\limits_0^1 \left( 3 x^2 + 2x + k \right) dx = 0,\] find the value of k.
If \[f\left( x \right) = \int_0^x t\sin tdt\], the write the value of \[f'\left( x \right)\]
The value of \[\int\limits_0^{2\pi} \sqrt{1 + \sin\frac{x}{2}}dx\] is
\[\int\limits_0^\infty \frac{1}{1 + e^x} dx\] equals
The value of \[\int\limits_0^{\pi/2} \cos x\ e^{\sin x}\ dx\] is
If f (a + b − x) = f (x), then \[\int\limits_a^b\] x f (x) dx is equal to
Evaluate: \[\int\limits_{- \pi/2}^{\pi/2} \frac{\cos x}{1 + e^x}dx\] .
\[\int\limits_0^4 x\sqrt{4 - x} dx\]
\[\int\limits_0^1 \tan^{- 1} x dx\]
\[\int\limits_0^{\pi/2} \frac{\sin^2 x}{\left( 1 + \cos x \right)^2} dx\]
\[\int\limits_0^{\pi/2} \frac{\sin x}{\sqrt{1 + \cos x}} dx\]
\[\int\limits_0^\infty \frac{x}{\left( 1 + x \right)\left( 1 + x^2 \right)} dx\]
\[\int\limits_0^{\pi/4} \sin 2x \sin 3x dx\]
\[\int\limits_0^{\pi/2} \frac{x}{\sin^2 x + \cos^2 x} dx\]
\[\int\limits_0^\pi \frac{dx}{6 - \cos x}dx\]
Using second fundamental theorem, evaluate the following:
`int_0^1 x"e"^(x^2) "d"x`
Using second fundamental theorem, evaluate the following:
`int_1^2 (x - 1)/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
Evaluate the following using properties of definite integral:
`int_0^1 log (1/x - 1) "d"x`
Choose the correct alternative:
Γ(1) is
Evaluate `int "dx"/sqrt((x - alpha)(beta - x)), beta > alpha`
Evaluate: `int_(-1)^2 |x^3 - 3x^2 + 2x|dx`
