Advertisements
Advertisements
प्रश्न
\[\int\limits_{\pi/6}^{\pi/2} \frac{\ cosec x \cot x}{1 + {cosec}^2 x} dx\]
उत्तर
\[\int_\frac{\pi}{6}^\frac{\pi}{2} \frac{\ cosecx\ cotx}{1 + \ cosec^2 x} d x\]
\[ = \int_\frac{\pi}{6}^\frac{\pi}{2} \frac{\ cosx}{1 + \sin^2 x} d x\]
\[ = \left[ \tan^{- 1} \left(\ sinx \right) \right]_\frac{\pi}{6}^\frac{\pi}{2} \]
\[ = \tan^{- 1} 1 - \tan^{- 1} \frac{1}{2}\]
\[ = \tan^{- 1} \frac{1 - \frac{1}{2}}{1 + 1 \times \frac{1}{2}}\]
\[ = \tan^{- 1} \frac{1}{3}\]
APPEARS IN
संबंधित प्रश्न
Evaluate each of the following integral:
Evaluate the following integral:
If f(2a − x) = −f(x), prove that
\[\int\limits_0^1 \left\{ x \right\} dx,\] where {x} denotes the fractional part of x.
The value of the integral \[\int\limits_0^{\pi/2} \frac{\sqrt{\cos x}}{\sqrt{\cos x} + \sqrt{\sin x}} dx\] is
\[\int\limits_0^1 \cos^{- 1} \left( \frac{1 - x^2}{1 + x^2} \right) dx\]
\[\int\limits_0^{\pi/4} \cos^4 x \sin^3 x dx\]
\[\int\limits_0^{\pi/2} \left| \sin x - \cos x \right| dx\]
\[\int\limits_1^3 \left| x^2 - 4 \right| dx\]
\[\int\limits_0^\pi \frac{x}{a^2 \cos^2 x + b^2 \sin^2 x} dx\]
\[\int\limits_0^{\pi/2} \frac{\sin^2 x}{\sin x + \cos x} dx\]
Using second fundamental theorem, evaluate the following:
`int_(-1)^1 (2x + 3)/(x^2 + 3x + 7) "d"x`
Evaluate the following using properties of definite integral:
`int_(- pi/4)^(pi/4) x^3 cos^3 x "d"x`
Evaluate the following using properties of definite integral:
`int_0^1 x/((1 - x)^(3/4)) "d"x`
Evaluate the following integrals as the limit of the sum:
`int_1^3 x "d"x`
Choose the correct alternative:
The value of `int_(- pi/2)^(pi/2) cos x "d"x` is
`int "e"^x ((1 - x)/(1 + x^2))^2 "d"x` is equal to ______.
