Advertisements
Advertisements
प्रश्न
विकल्प
0
π
π/2
π/4
उत्तर
π/2
\[\text{We have}, \]
\[I = \int_0^1 \frac{d}{dx}\left\{ \sin^{- 1} \left( \frac{2x}{1 + x^2} \right) \right\}dx\]
\[\text{We know since} \int f'(x) = f(x)\]
\[f(x) = si n^{- 1} \left( \frac{2x}{1 + x^2} \right) and f'(x) = \frac{d}{dx}\left\{ si n^{- 1} \left( \frac{2x}{1 + x^2} \right) \right\} \]
\[\text{Therefore}, I = \left[ \sin^{- 1} \left( \frac{2x}{1 + x^2} \right) \right]_0^1 \]
\[ = \sin^{- 1} \left( 1 \right) - \sin^{- 1} \left( 0 \right)\]
\[ = \frac{\pi}{2}\]
APPEARS IN
संबंधित प्रश्न
If \[f\left( a + b - x \right) = f\left( x \right)\] , then prove that \[\int_a^b xf\left( x \right)dx = \frac{a + b}{2} \int_a^b f\left( x \right)dx\]
If `f` is an integrable function such that f(2a − x) = f(x), then prove that
Evaluate each of the following integral:
If \[f\left( x \right) = \int_0^x t\sin tdt\], the write the value of \[f'\left( x \right)\]
\[\int\limits_0^1 \cos^{- 1} \left( \frac{1 - x^2}{1 + x^2} \right) dx\]
\[\int\limits_1^2 \frac{1}{x^2} e^{- 1/x} dx\]
\[\int\limits_1^2 \frac{x + 3}{x\left( x + 2 \right)} dx\]
\[\int\limits_{- 1/2}^{1/2} \cos x \log\left( \frac{1 + x}{1 - x} \right) dx\]
\[\int\limits_0^\pi \frac{x}{a^2 - \cos^2 x} dx, a > 1\]
Evaluate the following using properties of definite integral:
`int_(-1)^1 log ((2 - x)/(2 + x)) "d"x`
Evaluate the following:
`int_0^oo "e"^(-4x) x^4 "d"x`
Choose the correct alternative:
`int_0^1 (2x + 1) "d"x` is
Choose the correct alternative:
`int_0^oo x^4"e"^-x "d"x` is
Evaluate the following:
`int ((x^2 + 2))/(x + 1) "d"x`
