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
Evaluate the following integral:
Solve each of the following integral:
If \[I_{10} = \int\limits_0^{\pi/2} x^{10} \sin x\ dx,\] then the value of I10 + 90I8 is
The value of \[\int\limits_0^{\pi/2} \log\left( \frac{4 + 3 \sin x}{4 + 3 \cos x} \right) dx\] is
The value of \[\int\limits_{- \pi/2}^{\pi/2} \left( x^3 + x \cos x + \tan^5 x + 1 \right) dx, \] is
\[\int\limits_0^{\pi/2} \frac{\sin x}{\sqrt{1 + \cos x}} dx\]
\[\int\limits_0^{\pi/2} x^2 \cos 2x dx\]
\[\int\limits_0^1 \left| 2x - 1 \right| dx\]
\[\int\limits_0^\pi x \sin x \cos^4 x dx\]
\[\int\limits_0^{15} \left[ x^2 \right] dx\]
\[\int\limits_2^3 \frac{\sqrt{x}}{\sqrt{5 - x} + \sqrt{x}} dx\]
\[\int\limits_0^{\pi/2} \frac{1}{2 \cos x + 4 \sin x} dx\]
Evaluate the following:
`int_1^4` f(x) dx where f(x) = `{{:(4x + 3",", 1 ≤ x ≤ 2),(3x + 5",", 2 < x ≤ 4):}`
Evaluate the following integrals as the limit of the sum:
`int_0^1 (x + 4) "d"x`
Evaluate the following integrals as the limit of the sum:
`int_1^3 (2x + 3) "d"x`
Evaluate `int "dx"/sqrt((x - alpha)(beta - x)), beta > alpha`
If `int (3"e"^x - 5"e"^-x)/(4"e"6x + 5"e"^-x)"d"x` = ax + b log |4ex + 5e –x| + C, then ______.
