Advertisements
Advertisements
Question
Solution
\[Let\ I = \int_0^4 \frac{1}{\sqrt{4x - x^2}} d x . Then, \]
\[I = \int_0^4 \frac{1}{\sqrt{4x - x^2 - 4 + 4}} d x\]
\[ \Rightarrow I = \int_0^4 \frac{1}{\sqrt{- \left( x - 2 \right)^2 + 4}} d x\]
\[ \Rightarrow I = \left[ \sin^{- 1} \frac{\left( x - 2 \right)}{2} \right]_0^4 \]
\[ \Rightarrow I = \left( \sin^{- 1} 1 - \sin^{- 1} ( - 1) \right)\]
\[ \Rightarrow I = 2 \sin^{- 1} 1\]
\[ \Rightarrow I = 2 \frac{\pi}{2} = \pi\]
APPEARS IN
RELATED QUESTIONS
Evaluate the following definite integrals:
Evaluate the following integral:
Evaluate each of the following integral:
Evaluate each of the following integral:
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
The value of \[\int\limits_0^{\pi/2} \cos x\ e^{\sin x}\ dx\] is
Evaluate : \[\int\limits_0^\pi/4 \frac{\sin x + \cos x}{16 + 9 \sin 2x}dx\] .
\[\int\limits_0^1 \left| 2x - 1 \right| dx\]
\[\int\limits_0^\pi \frac{x \sin x}{1 + \cos^2 x} dx\]
\[\int\limits_0^{\pi/2} \frac{x}{\sin^2 x + \cos^2 x} dx\]
\[\int\limits_0^3 \left( x^2 + 1 \right) dx\]
Using second fundamental theorem, evaluate the following:
`int_1^"e" ("d"x)/(x(1 + logx)^3`
Evaluate the following using properties of definite integral:
`int_0^(i/2) (sin^7x)/(sin^7x + cos^7x) "d"x`
Evaluate the following:
`Γ (9/2)`
Evaluate the following:
`int_0^oo "e"^(-mx) x^6 "d"x`
Evaluate `int (x^2"d"x)/(x^4 + x^2 - 2)`
`int (x + 3)/(x + 4)^2 "e"^x "d"x` = ______.