Advertisements
Advertisements
Question
Advertisements
Solution
\[\int_0^1 \tan^{- 1} \left( \frac{2x}{1 - x^2} \right) d x\]
\[ = \int_0^1 2 \tan^{- 1} x\]
\[ = 2 \left[ x \tan^{- 1} x \right]_0^1 - 2 \int_0^1 \frac{x}{1 + x^2}dx\]
\[ = 2 \left[ x \tan^{- 1} x \right]_0^1 - \left[ \log\left( 1 + x^2 \right) \right]_0^1 \]
\[ = 2\frac{\pi}{4} - 0 - \log2 + 0\]
\[ = \frac{\pi}{2} - \log2\]
APPEARS IN
RELATED QUESTIONS
\[\int\limits_{\pi/4}^{\pi/2} \cot x\ dx\]
\[\int\limits_0^\pi \frac{1}{1 + \sin x} dx\] equals
The value of \[\int\limits_0^\pi \frac{x \tan x}{\sec x + \cos x} dx\] is __________ .
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 value of \[\int\limits_{- \pi}^\pi \sin^3 x \cos^2 x\ dx\] is
Evaluate : \[\int\limits_0^\pi/4 \frac{\sin x + \cos x}{16 + 9 \sin 2x}dx\] .
`int_0^(2a)f(x)dx`
\[\int\limits_0^{\pi/2} \frac{\sin x}{\sqrt{1 + \cos x}} dx\]
Using second fundamental theorem, evaluate the following:
`int_0^3 ("e"^x "d"x)/(1 + "e"^x)`
Choose the correct alternative:
If n > 0, then Γ(n) is
