Advertisements
Advertisements
प्रश्न
Advertisements
उत्तर
\[Let\ I = \int_0^1 \frac{\tan^{- 1} x}{1 + x^2} d\ x . Then, \]
\[Let\ \tan^{- 1} x = t . Then, \frac{1}{1 + x^2} dx = dt\]
\[When\ x = 0, t = 0\ and\ x = 1, t = \frac{\pi}{4}\]
\[ \therefore I = \int_0^\frac{\pi}{4} t dt\]
\[ \Rightarrow I = \left[ \frac{t^2}{2} \right]_0^\frac{\pi}{4} \]
\[ \Rightarrow I = \frac{\pi^2}{32}\]
APPEARS IN
संबंधित प्रश्न
\[\int\limits_1^4 f\left( x \right) dx, where f\left( x \right) = \begin{cases}7x + 3 & , & \text{if }1 \leq x \leq 3 \\ 8x & , & \text{if }3 \leq x \leq 4\end{cases}\]
If `f` is an integrable function such that f(2a − x) = f(x), then prove that
If f(x) is a continuous function defined on [−a, a], then prove that
\[\int\limits_0^1 \left\{ x \right\} dx,\] where {x} denotes the fractional part of x.
\[\int\limits_0^\pi \frac{1}{1 + \sin x} dx\] equals
`int_0^1 sqrt((1 - "x")/(1 + "x")) "dx"`
The value of \[\int\limits_0^\pi \frac{1}{5 + 3 \cos x} dx\] is
Evaluate : \[\int\limits_0^{2\pi} \cos^5 x dx\] .
\[\int\limits_0^1 \cos^{- 1} x dx\]
\[\int\limits_0^1 \cos^{- 1} \left( \frac{1 - x^2}{1 + x^2} \right) dx\]
\[\int\limits_1^2 \frac{1}{x^2} e^{- 1/x} dx\]
\[\int\limits_0^a \frac{\sqrt{x}}{\sqrt{x} + \sqrt{a - x}} dx\]
\[\int\limits_0^\pi x \sin x \cos^4 x dx\]
\[\int\limits_0^{\pi/2} \frac{x \sin x \cos x}{\sin^4 x + \cos^4 x} dx\]
\[\int\limits_0^\pi \frac{x \tan x}{\sec x + \tan x} dx\]
\[\int\limits_0^1 \cot^{- 1} \left( 1 - x + x^2 \right) dx\]
Evaluate the following using properties of definite integral:
`int_0^(i/2) (sin^7x)/(sin^7x + cos^7x) "d"x`
Evaluate `int sqrt((1 + x)/(1 - x)) "d"x`, x ≠1
