Advertisements
Advertisements
प्रश्न
पर्याय
loge 3
- \[\log_e \sqrt{3}\]
- \[\frac{1}{2}\log\left( - 1 \right)\]
log (−1)
Advertisements
उत्तर
\[\int_\frac{\pi}{6}^\frac{\pi}{3} \frac{1}{\sin2x} d x\]
\[ = \int_\frac{\pi}{6}^\frac{\pi}{3} \ cosec2x\ dx\]
\[ = \frac{1}{2} \int_\frac{\pi}{6}^\frac{\pi}{3} 2\ cosec2x\ dx\]
\[ = \frac{- 1}{2} \left[ \log\left( \ cosec\ 2x\ + \cot2x \right) \right]_\frac{\pi}{6}^\frac{\pi}{3} \]
\[ = \frac{- 1}{2}\left[ - 2\log\sqrt{3} \right]\]
\[ = \log\sqrt{3}\]
APPEARS IN
संबंधित प्रश्न
Evaluate each of the following integral:
If f is an integrable function, show that
\[\int\limits_{- a}^a f\left( x^2 \right) dx = 2 \int\limits_0^a f\left( x^2 \right) dx\]
\[\int\limits_0^\infty \frac{1}{1 + e^x} dx\] equals
Evaluate : \[\int\limits_0^\pi \frac{x}{1 + \sin \alpha \sin x}dx\] .
`int_0^(2a)f(x)dx`
\[\int\limits_0^1 \cos^{- 1} x dx\]
\[\int\limits_0^1 \tan^{- 1} \left( \frac{2x}{1 - x^2} \right) dx\]
\[\int\limits_0^{\pi/2} \frac{\sin^2 x}{\left( 1 + \cos x \right)^2} dx\]
\[\int\limits_0^{\pi/2} x^2 \cos 2x dx\]
\[\int\limits_1^2 \frac{x + 3}{x\left( x + 2 \right)} dx\]
\[\int\limits_0^{\pi/4} \tan^4 x dx\]
\[\int\limits_0^{\pi/2} \frac{x \sin x \cos x}{\sin^4 x + \cos^4 x} dx\]
\[\int\limits_0^3 \left( x^2 + 1 \right) dx\]
Using second fundamental theorem, evaluate the following:
`int_0^(1/4) sqrt(1 - 4) "d"x`
Evaluate the following:
f(x) = `{{:("c"x",", 0 < x < 1),(0",", "otherwise"):}` Find 'c" if `int_0^1 "f"(x) "d"x` = 2
Evaluate the following using properties of definite integral:
`int_(- pi/2)^(pi/2) sin^2theta "d"theta`
Evaluate the following using properties of definite integral:
`int_(-1)^1 log ((2 - x)/(2 + x)) "d"x`
