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
Evaluate each of the following integral:
Evaluate each of the following integral:
`int_0^1 sqrt((1 - "x")/(1 + "x")) "dx"`
\[\int\limits_0^4 x\sqrt{4 - x} dx\]
\[\int\limits_1^5 \frac{x}{\sqrt{2x - 1}} dx\]
\[\int\limits_1^2 \frac{1}{x^2} e^{- 1/x} dx\]
\[\int\limits_0^1 \log\left( 1 + x \right) dx\]
\[\int\limits_{- 1/2}^{1/2} \cos x \log\left( \frac{1 + x}{1 - x} \right) dx\]
\[\int\limits_0^\pi \frac{x}{1 + \cos \alpha \sin x} dx\]
\[\int\limits_0^\pi \frac{dx}{6 - \cos x}dx\]
\[\int\limits_0^{\pi/2} \frac{1}{2 \cos x + 4 \sin x} dx\]
Choose the correct alternative:
If f(x) is a continuous function and a < c < b, then `int_"a"^"c" f(x) "d"x + int_"c"^"b" f(x) "d"x` is
Choose the correct alternative:
The value of `int_(- pi/2)^(pi/2) cos x "d"x` is
Choose the correct alternative:
Using the factorial representation of the gamma function, which of the following is the solution for the gamma function Γ(n) when n = 8 is
Choose the correct alternative:
If n > 0, then Γ(n) is
Evaluate `int (3"a"x)/("b"^2 + "c"^2x^2) "d"x`
