Advertisements
Advertisements
प्रश्न
\[\int\limits_0^1 \cos^{- 1} x dx\]
Advertisements
उत्तर
\[\int_0^1 \cos^{- 1} x d x\]
\[ = \int_0^1 \left( \cos^{- 1} x \times 1 \right) d x\]
\[ = \left[ \cos^{- 1} x x \right]_0^1 - \int_0^1 \frac{- x}{\sqrt{1 - x^2}}dx\]
\[ = \left[ x \cos^{- 1} x \right]_0^1 - \frac{2}{2} \left[ \sqrt{1 - x^2} \right]_0^1 \]
\[ = 0 + 1\]
\[ = 1\]
APPEARS IN
संबंधित प्रश्न
\[\int\limits_0^{( \pi )^{2/3}} \sqrt{x} \cos^2 x^{3/2} dx\]
If f is an integrable function, show that
\[\int\limits_{- a}^a f\left( x^2 \right) dx = 2 \int\limits_0^a f\left( x^2 \right) dx\]
If f(x) is a continuous function defined on [−a, a], then prove that
If \[\int\limits_0^1 \left( 3 x^2 + 2x + k \right) dx = 0,\] find the value of k.
Given that \[\int\limits_0^\infty \frac{x^2}{\left( x^2 + a^2 \right)\left( x^2 + b^2 \right)\left( x^2 + c^2 \right)} dx = \frac{\pi}{2\left( a + b \right)\left( b + c \right)\left( c + a \right)},\] the value of \[\int\limits_0^\infty \frac{dx}{\left( x^2 + 4 \right)\left( x^2 + 9 \right)},\]
The value of \[\int\limits_{- \pi/2}^{\pi/2} \left( x^3 + x \cos x + \tan^5 x + 1 \right) dx, \] is
Evaluate : \[\int\frac{dx}{\sin^2 x \cos^2 x}\] .
\[\int\limits_0^{\pi/3} \frac{\cos x}{3 + 4 \sin x} dx\]
\[\int\limits_0^{\pi/4} e^x \sin x dx\]
\[\int\limits_0^1 \left| \sin 2\pi x \right| dx\]
\[\int\limits_0^\pi \frac{x}{a^2 \cos^2 x + b^2 \sin^2 x} dx\]
\[\int\limits_0^\pi \frac{dx}{6 - \cos x}dx\]
\[\int\limits_0^2 \left( x^2 + 2 \right) dx\]
Find : `∫_a^b logx/x` dx
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 x/((1 - x)^(3/4)) "d"x`
If `int (3"e"^x - 5"e"^-x)/(4"e"6x + 5"e"^-x)"d"x` = ax + b log |4ex + 5e –x| + C, then ______.
