Advertisements
Advertisements
प्रश्न
Advertisements
उत्तर
\[Let\ I = \int_0^\frac{\pi}{2} \frac{\sin \theta}{\sqrt{1 + \cos \theta}} d \theta . \]
\[Let\ \cos\ \theta = t . Then, - \sin\ \theta\ d\theta\ = dt\]
\[When\ \theta = 0, t = 1\ and\ \theta = \frac{\pi}{2}, t = 0\]
\[ \therefore I = \int_0^\frac{\pi}{2} \frac{\sin \theta}{\sqrt{1 + \cos \theta}} d \theta\]
\[ = \int_1^0 \frac{- dt}{\sqrt{1 + t}}\]
\[ = \int_0^1 \frac{dt}{\sqrt{1 + t}}\]
\[ = 2 \left[ \sqrt{1 + t} \right]_0^1 \]
\[ = 2\left( \sqrt{2} - 1 \right)\]
APPEARS IN
संबंधित प्रश्न
\[\int\limits_0^{( \pi )^{2/3}} \sqrt{x} \cos^2 x^{3/2} dx\]
Evaluate the following integral:
The value of \[\int\limits_{- \pi/2}^{\pi/2} \left( x^3 + x \cos x + \tan^5 x + 1 \right) dx, \] is
\[\int\limits_0^4 x\sqrt{4 - x} dx\]
\[\int\limits_1^5 \frac{x}{\sqrt{2x - 1}} dx\]
\[\int\limits_0^{\pi/2} \frac{\sin x}{\sqrt{1 + \cos x}} dx\]
\[\int\limits_0^{\pi/2} \frac{\cos x}{1 + \sin^2 x} dx\]
\[\int\limits_0^1 \log\left( 1 + x \right) dx\]
\[\int\limits_0^1 \left( \cos^{- 1} x \right)^2 dx\]
\[\int\limits_0^\pi \frac{x \tan x}{\sec x + \tan x} dx\]
\[\int\limits_{- 1}^1 e^{2x} dx\]
Using second fundamental theorem, evaluate the following:
`int_1^2 (x - 1)/x^2 "d"x`
Evaluate the following:
`int_0^oo "e"^(-mx) x^6 "d"x`
Evaluate the following integrals as the limit of the sum:
`int_0^1 x^2 "d"x`
Choose the correct alternative:
`int_(-1)^1 x^3 "e"^(x^4) "d"x` is
Evaluate `int sqrt((1 + x)/(1 - x)) "d"x`, x ≠1
Find `int sqrt(10 - 4x + 4x^2) "d"x`
