Advertisements
Advertisements
प्रश्न
Advertisements
उत्तर
\[Let\ I = \int_{- 1}^1 \frac{1}{1 + x^2} d x . Then, \]
\[I = \left[ \tan^{- 1} x \right]_{- 1}^1 \]
\[ \Rightarrow I = \tan^{- 1} 1 - \tan^{- 1} \left( - 1 \right)\]
\[ \Rightarrow I = \frac{\pi}{4} - \left( - \frac{\pi}{4} \right)\]
\[ \Rightarrow I = \frac{\pi}{2}\]
APPEARS IN
संबंधित प्रश्न
Evaluate the following integral:
If f (a + b − x) = f (x), then \[\int\limits_a^b\] x f (x) dx is equal to
Evaluate : \[\int\frac{dx}{\sin^2 x \cos^2 x}\] .
\[\int\limits_0^\pi \sin^3 x\left( 1 + 2 \cos x \right) \left( 1 + \cos x \right)^2 dx\]
\[\int\limits_0^{\pi/4} \cos^4 x \sin^3 x dx\]
\[\int\limits_{\pi/3}^{\pi/2} \frac{\sqrt{1 + \cos x}}{\left( 1 - \cos x \right)^{5/2}} dx\]
\[\int\limits_0^1 \cot^{- 1} \left( 1 - x + x^2 \right) dx\]
\[\int\limits_0^{\pi/2} \frac{dx}{4 \cos x + 2 \sin x}dx\]
Using second fundamental theorem, evaluate the following:
`int_1^2 (x "d"x)/(x^2 + 1)`
If f(x) = `{{:(x^2"e"^(-2x)",", x ≥ 0),(0",", "otherwise"):}`, then evaluate `int_0^oo "f"(x) "d"x`
Choose the correct alternative:
`int_0^1 (2x + 1) "d"x` is
Choose the correct alternative:
Γ(1) is
Choose the correct alternative:
`Γ(3/2)`
If `int (3"e"^x - 5"e"^-x)/(4"e"6x + 5"e"^-x)"d"x` = ax + b log |4ex + 5e –x| + C, then ______.
Verify the following:
`int (x - 1)/(2x + 3) "d"x = x - log |(2x + 3)^2| + "C"`
