Advertisements
Advertisements
Question
Advertisements
Solution
\[Let\ I = \int_0^\frac{\pi}{2} \frac{\sin \theta}{\sqrt{1 + \cos \theta}} d \theta . \]
\[Let\ \cos\ \theta = t . Then, - \sin\ \theta\ d\theta\ = dt\]
\[When\ \theta = 0, t = 1\ and\ \theta = \frac{\pi}{2}, t = 0\]
\[ \therefore I = \int_0^\frac{\pi}{2} \frac{\sin \theta}{\sqrt{1 + \cos \theta}} d \theta\]
\[ = \int_1^0 \frac{- dt}{\sqrt{1 + t}}\]
\[ = \int_0^1 \frac{dt}{\sqrt{1 + t}}\]
\[ = 2 \left[ \sqrt{1 + t} \right]_0^1 \]
\[ = 2\left( \sqrt{2} - 1 \right)\]
APPEARS IN
RELATED QUESTIONS
\[\int\limits_0^{( \pi )^{2/3}} \sqrt{x} \cos^2 x^{3/2} dx\]
Evaluate the following integral:
If \[\int\limits_0^a 3 x^2 dx = 8,\] write the value of a.
\[\int\limits_0^{\pi/2} \frac{1}{2 + \cos x} dx\] equals
\[\int\limits_0^{2\pi} \cos^7 x dx\]
\[\int\limits_0^{\pi/2} \frac{x \sin x \cos x}{\sin^4 x + \cos^4 x} dx\]
\[\int\limits_0^\pi \frac{x \tan x}{\sec x + \tan x} dx\]
\[\int\limits_1^4 \left( x^2 + x \right) dx\]
\[\int\limits_0^3 \left( x^2 + 1 \right) dx\]
Find : `∫_a^b logx/x` dx
Using second fundamental theorem, evaluate the following:
`int_0^1 "e"^(2x) "d"x`
Evaluate the following:
`int_0^oo "e"^(-4x) x^4 "d"x`
Find `int x^2/(x^4 + 3x^2 + 2) "d"x`
Find `int sqrt(10 - 4x + 4x^2) "d"x`
If `intx^3/sqrt(1 + x^2) "d"x = "a"(1 + x^2)^(3/2) + "b"sqrt(1 + x^2) + "C"`, then ______.
