Advertisements
Advertisements
प्रश्न
Advertisements
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
If f is an integrable function, show that
Evaluate each of the following integral:
Evaluate each of the following integral:
The value of \[\int\limits_0^\pi \frac{1}{5 + 3 \cos x} dx\] is
The value of \[\int\limits_0^{\pi/2} \log\left( \frac{4 + 3 \sin x}{4 + 3 \cos x} \right) dx\] is
Evaluate : \[\int\limits_0^\pi \frac{x}{1 + \sin \alpha \sin x}dx\] .
\[\int\limits_0^\pi \sin^3 x\left( 1 + 2 \cos x \right) \left( 1 + \cos x \right)^2 dx\]
\[\int\limits_{\pi/3}^{\pi/2} \frac{\sqrt{1 + \cos x}}{\left( 1 - \cos x \right)^{5/2}} dx\]
\[\int\limits_0^{\pi/2} \left| \sin x - \cos x \right| dx\]
\[\int\limits_0^\pi \frac{x \sin x}{1 + \cos^2 x} dx\]
\[\int\limits_{- \pi/4}^{\pi/4} \left| \tan x \right| dx\]
Evaluate the following:
`int_1^4` f(x) dx where f(x) = `{{:(4x + 3",", 1 ≤ x ≤ 2),(3x + 5",", 2 < x ≤ 4):}`
Evaluate the following integrals as the limit of the sum:
`int_0^1 x^2 "d"x`
Choose the correct alternative:
`int_0^1 (2x + 1) "d"x` is
Verify the following:
`int (2x + 3)/(x^2 + 3x) "d"x = log|x^2 + 3x| + "C"`
