Advertisements
Advertisements
Question
\[\int\limits_0^1 \cos^{- 1} x dx\]
Advertisements
Solution
\[\int_0^1 \cos^{- 1} x d x\]
\[ = \int_0^1 \left( \cos^{- 1} x \times 1 \right) d x\]
\[ = \left[ \cos^{- 1} x x \right]_0^1 - \int_0^1 \frac{- x}{\sqrt{1 - x^2}}dx\]
\[ = \left[ x \cos^{- 1} x \right]_0^1 - \frac{2}{2} \left[ \sqrt{1 - x^2} \right]_0^1 \]
\[ = 0 + 1\]
\[ = 1\]
APPEARS IN
RELATED QUESTIONS
The value of \[\int\limits_0^\pi \frac{x \tan x}{\sec x + \cos x} dx\] is __________ .
\[\int_0^\frac{\pi^2}{4} \frac{\sin\sqrt{x}}{\sqrt{x}} dx\] equals
Evaluate : \[\int\limits_0^\pi \frac{x}{1 + \sin \alpha \sin x}dx\] .
Evaluate: \[\int\limits_{- \pi/2}^{\pi/2} \frac{\cos x}{1 + e^x}dx\] .
\[\int\limits_0^{\pi/4} \sin 2x \sin 3x dx\]
\[\int\limits_0^1 \sqrt{\frac{1 - x}{1 + x}} dx\]
\[\int\limits_0^1 \log\left( 1 + x \right) dx\]
\[\int\limits_0^{\pi/4} \tan^4 x dx\]
\[\int\limits_0^1 \left| 2x - 1 \right| dx\]
\[\int\limits_0^{2\pi} \cos^7 x dx\]
\[\int\limits_0^\pi \frac{x}{a^2 \cos^2 x + b^2 \sin^2 x} dx\]
\[\int\limits_2^3 e^{- x} dx\]
Find : `∫_a^b logx/x` dx
Using second fundamental theorem, evaluate the following:
`int_0^1 "e"^(2x) "d"x`
Choose the correct alternative:
If f(x) is a continuous function and a < c < b, then `int_"a"^"c" f(x) "d"x + int_"c"^"b" f(x) "d"x` is
Choose the correct alternative:
Using the factorial representation of the gamma function, which of the following is the solution for the gamma function Γ(n) when n = 8 is
`int (cos2x - cos 2theta)/(cosx - costheta) "d"x` is equal to ______.
