Advertisements
Advertisements
Question
Options
loge 3
- \[\log_e \sqrt{3}\]
- \[\frac{1}{2}\log\left( - 1 \right)\]
log (−1)
Advertisements
Solution
\[\int_\frac{\pi}{6}^\frac{\pi}{3} \frac{1}{\sin2x} d x\]
\[ = \int_\frac{\pi}{6}^\frac{\pi}{3} \ cosec2x\ dx\]
\[ = \frac{1}{2} \int_\frac{\pi}{6}^\frac{\pi}{3} 2\ cosec2x\ dx\]
\[ = \frac{- 1}{2} \left[ \log\left( \ cosec\ 2x\ + \cot2x \right) \right]_\frac{\pi}{6}^\frac{\pi}{3} \]
\[ = \frac{- 1}{2}\left[ - 2\log\sqrt{3} \right]\]
\[ = \log\sqrt{3}\]
APPEARS IN
RELATED QUESTIONS
Evaluate the following integral:
Evaluate the following integral:
If `f` is an integrable function such that f(2a − x) = f(x), then prove that
Evaluate each of the following integral:
\[\int\limits_0^{2a} f\left( x \right) dx\] is equal to
The value of \[\int\limits_{- \pi/2}^{\pi/2} \left( x^3 + x \cos x + \tan^5 x + 1 \right) dx, \] is
\[\int\limits_0^{1/\sqrt{3}} \tan^{- 1} \left( \frac{3x - x^3}{1 - 3 x^2} \right) dx\]
\[\int\limits_0^{\pi/2} \frac{\cos x}{1 + \sin^2 x} dx\]
\[\int\limits_0^1 \sqrt{\frac{1 - x}{1 + x}} dx\]
\[\int\limits_0^1 \log\left( 1 + x \right) dx\]
\[\int\limits_{- 1/2}^{1/2} \cos x \log\left( \frac{1 + x}{1 - x} \right) dx\]
\[\int\limits_0^\pi \frac{dx}{6 - \cos x}dx\]
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 using properties of definite integral:
`int_(- pi/4)^(pi/4) x^3 cos^3 x "d"x`
Evaluate the following:
`int_0^oo "e"^(-4x) x^4 "d"x`
Integrate `((2"a")/sqrt(x) - "b"/x^2 + 3"c"root(3)(x^2))` w.r.t. x
`int (cos2x - cos 2theta)/(cosx - costheta) "d"x` is equal to ______.
`int x^3/(x + 1)` is equal to ______.
