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
\[\int\limits_1^4 f\left( x \right) dx, where f\left( x \right) = \begin{cases}7x + 3 & , & \text{if }1 \leq x \leq 3 \\ 8x & , & \text{if }3 \leq x \leq 4\end{cases}\]
Prove that:
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 \[\int\limits_0^a 3 x^2 dx = 8,\] write the value of a.
\[\int\limits_0^\pi \frac{1}{1 + \sin x} dx\] equals
\[\int_0^\frac{\pi^2}{4} \frac{\sin\sqrt{x}}{\sqrt{x}} dx\] equals
The value of the integral \[\int\limits_{- 2}^2 \left| 1 - x^2 \right| dx\] is ________ .
Evaluate the following integrals :-
\[\int_2^4 \frac{x^2 + x}{\sqrt{2x + 1}}dx\]
\[\int\limits_{- a}^a \frac{x e^{x^2}}{1 + x^2} dx\]
\[\int\limits_0^{\pi/2} \frac{1}{1 + \cot^7 x} dx\]
\[\int\limits_0^\pi \frac{x \tan x}{\sec x + \tan x} dx\]
\[\int\limits_2^3 \frac{\sqrt{x}}{\sqrt{5 - x} + \sqrt{x}} dx\]
\[\int\limits_0^\pi \frac{dx}{6 - \cos x}dx\]
\[\int\limits_0^2 \left( 2 x^2 + 3 \right) dx\]
Using second fundamental theorem, evaluate the following:
`int_0^1 "e"^(2x) "d"x`
Using second fundamental theorem, evaluate the following:
`int_0^(pi/2) sqrt(1 + cos x) "d"x`
Evaluate the following using properties of definite integral:
`int_(- pi/2)^(pi/2) sin^2theta "d"theta`
Choose the correct alternative:
Γ(1) is
Evaluate: `int_(-1)^2 |x^3 - 3x^2 + 2x|dx`
