Advertisements
Advertisements
प्रश्न
पर्याय
π
π/2
0
2π
Advertisements
उत्तर
0
\[I = \int_0^\frac{\pi}{2} \sin2x \log \tan x\ d x . . . . . \left( 1 \right)\]
\[I = \int_0^\frac{\pi}{2} \sin\left( \pi - 2x \right) \log \tan\left( \frac{\pi}{2} - x \right) d x\]
\[I = \int_0^\frac{\pi}{2} \sin2x \log \cot x\ d x . . . . . \left( 2 \right)\]
\[\text{Adding} \left( 1 \right) and \left( 2 \right), \text{we get}, \]
\[2I = \int_0^\frac{\pi}{2} \sin2x\left( \log \tan x + \log \cot x \right) d x\]
\[2I = \int_0^\frac{\pi}{2} \sin2x\left( \log \tan x \cot x \right) d x\]
\[2I = \int_0^\frac{\pi}{2} \sin2x\left( \log1 \right) d x\]
\[I = 0\]
APPEARS IN
संबंधित प्रश्न
\[\int\limits_1^4 f\left( x \right) dx, where f\left( x \right) = \begin{cases}7x + 3 & , & \text{if }1 \leq x \leq 3 \\ 8x & , & \text{if }3 \leq x \leq 4\end{cases}\]
The value of \[\int\limits_0^{\pi/2} \cos x\ e^{\sin x}\ dx\] is
The derivative of \[f\left( x \right) = \int\limits_{x^2}^{x^3} \frac{1}{\log_e t} dt, \left( x > 0 \right),\] is
The value of the integral \[\int\limits_{- 2}^2 \left| 1 - x^2 \right| dx\] is ________ .
Evaluate : \[\int\limits_0^\pi/4 \frac{\sin x + \cos x}{16 + 9 \sin 2x}dx\] .
\[\int\limits_0^1 \cos^{- 1} \left( \frac{1 - x^2}{1 + x^2} \right) dx\]
\[\int\limits_0^{\pi/2} \frac{\cos x}{1 + \sin^2 x} dx\]
\[\int\limits_0^\pi x \sin x \cos^4 x dx\]
\[\int\limits_0^{15} \left[ x^2 \right] dx\]
\[\int\limits_0^1 \cot^{- 1} \left( 1 - x + x^2 \right) dx\]
\[\int\limits_0^{\pi/2} \frac{dx}{4 \cos x + 2 \sin x}dx\]
Using second fundamental theorem, evaluate the following:
`int_0^1 x"e"^(x^2) "d"x`
Choose the correct alternative:
If n > 0, then Γ(n) is
Evaluate the following:
`int ((x^2 + 2))/(x + 1) "d"x`
`int "e"^x ((1 - x)/(1 + x^2))^2 "d"x` is equal to ______.
If `intx^3/sqrt(1 + x^2) "d"x = "a"(1 + x^2)^(3/2) + "b"sqrt(1 + x^2) + "C"`, then ______.
