Advertisements
Advertisements
प्रश्न
\[\int\limits_0^1 \cos^{- 1} \left( \frac{1 - x^2}{1 + x^2} \right) dx\]
Advertisements
उत्तर
\[\int_0^1 \cos^{- 1} \left( \frac{1 - x^2}{1 + x^2} \right) d x\]
\[Let, x = \tan\theta, dx = se c^2 \theta d\theta\]
\[\text{When, }x \to 0 ; \theta \to 0\]
\[\text{and }x \to 1 ; \theta \to \frac{\pi}{4}\]
Therefore, the integral becomes
\[ \int_0^\frac{\pi}{4} \cos^{- 1} \left( \frac{1 - \tan^2 \theta}{1 + \tan^2 \theta} \right) se c^2 \theta d\theta\]
\[ = \int_0^\frac{\pi}{4} \cos^{- 1} \left( \cos2\theta \right) se c^2 \theta d\theta\]
\[ = 2 \int_0^\frac{\pi}{4} \theta se c^2 \theta d\theta\]
\[ = 2 \left[ \theta tan\theta \right]_0^\frac{\pi}{4} - 2 \int_0^\frac{\pi}{4} \tan\theta d\theta\]
\[ = 2 \left[ \theta \tan\ theta \right]_0^\frac{\pi}{4} + 2 \left[ \log\left( \cos\theta \right) \right]_0^\frac{\pi}{4} \]
\[ = 2\left( \frac{\pi}{4} - 0 \right) + 2\left[ \log\frac{1}{\sqrt{2}} - 0 \right]\]
\[ = \frac{\pi}{2} - \log2\]
APPEARS IN
संबंधित प्रश्न
If \[\int\limits_0^a 3 x^2 dx = 8,\] write the value of a.
If \[\int\limits_0^1 f\left( x \right) dx = 1, \int\limits_0^1 xf\left( x \right) dx = a, \int\limits_0^1 x^2 f\left( x \right) dx = a^2 , then \int\limits_0^1 \left( a - x \right)^2 f\left( x \right) dx\] equals
The derivative of \[f\left( x \right) = \int\limits_{x^2}^{x^3} \frac{1}{\log_e t} dt, \left( x > 0 \right),\] is
\[\int\limits_0^{2a} f\left( x \right) dx\] is equal to
The value of \[\int\limits_{- \pi/2}^{\pi/2} \left( x^3 + x \cos x + \tan^5 x + 1 \right) dx, \] is
\[\int\limits_1^2 \frac{1}{x^2} e^{- 1/x} dx\]
\[\int\limits_0^{\pi/4} e^x \sin x dx\]
\[\int\limits_1^3 \left| x^2 - 2x \right| dx\]
Choose the correct alternative:
`int_0^oo x^4"e"^-x "d"x` is
Evaluate `int (3"a"x)/("b"^2 + "c"^2x^2) "d"x`
Evaluate `int (x^2"d"x)/(x^4 + x^2 - 2)`
`int "e"^x ((1 - x)/(1 + x^2))^2 "d"x` is equal to ______.
Evaluate: `int_(-1)^2 |x^3 - 3x^2 + 2x|dx`
