Advertisements
Advertisements
प्रश्न
उत्तर
\[\int_0^1 \tan^{- 1} \left( \frac{2x}{1 - x^2} \right) d x\]
\[ = \int_0^1 2 \tan^{- 1} x\]
\[ = 2 \left[ x \tan^{- 1} x \right]_0^1 - 2 \int_0^1 \frac{x}{1 + x^2}dx\]
\[ = 2 \left[ x \tan^{- 1} x \right]_0^1 - \left[ \log\left( 1 + x^2 \right) \right]_0^1 \]
\[ = 2\frac{\pi}{4} - 0 - \log2 + 0\]
\[ = \frac{\pi}{2} - \log2\]
APPEARS IN
संबंधित प्रश्न
The value of \[\int\limits_0^{\pi/2} \cos x\ e^{\sin x}\ dx\] is
The value of \[\int\limits_0^{\pi/2} \log\left( \frac{4 + 3 \sin x}{4 + 3 \cos x} \right) dx\] is
Evaluate: \[\int\limits_{- \pi/2}^{\pi/2} \frac{\cos x}{1 + e^x}dx\] .
\[\int\limits_0^1 \log\left( 1 + x \right) dx\]
\[\int\limits_0^\pi x \sin x \cos^4 x dx\]
\[\int\limits_0^{15} \left[ x^2 \right] dx\]
\[\int\limits_0^1 \cot^{- 1} \left( 1 - x + x^2 \right) dx\]
\[\int\limits_{\pi/6}^{\pi/2} \frac{\ cosec x \cot x}{1 + {cosec}^2 x} dx\]
\[\int\limits_0^3 \left( x^2 + 1 \right) dx\]
Using second fundamental theorem, evaluate the following:
`int_1^2 (x "d"x)/(x^2 + 1)`
Using second fundamental theorem, evaluate the following:
`int_1^"e" ("d"x)/(x(1 + logx)^3`
Evaluate the following:
Γ(4)
Evaluate the following:
`Γ (9/2)`
If f(x) = `{{:(x^2"e"^(-2x)",", x ≥ 0),(0",", "otherwise"):}`, then evaluate `int_0^oo "f"(x) "d"x`
Evaluate the following integrals as the limit of the sum:
`int_1^3 (2x + 3) "d"x`
Evaluate `int (3"a"x)/("b"^2 + "c"^2x^2) "d"x`
`int x^9/(4x^2 + 1)^6 "d"x` is equal to ______.
Evaluate: `int_(-1)^2 |x^3 - 3x^2 + 2x|dx`
