Advertisements
Advertisements
Question
\[\int\limits_0^{\pi/2} \frac{\sin x}{\sqrt{1 + \cos x}} dx\]
Advertisements
Solution
\[\int_0^\frac{\pi}{2} \frac{\sin x}{\sqrt{1 + \cos x}} d x\]
\[Let 1 + \cos x = t, then - \sin x dx = dt\]
\[When, x \to 0, t \to 2 and x \to \frac{\pi}{2}, t \to 1\]
Therefore, the integral becomes
\[ \int_2^1 \frac{- 1}{\sqrt{t}}dt\]
\[ = \int_1^2 \frac{1}{\sqrt{t}}dt\]
\[ = 2 \left[ \sqrt{t} \right]_1^2 \]
\[ = 2\left( \sqrt{2} - 1 \right)\]
APPEARS IN
RELATED QUESTIONS
The value of the integral \[\int\limits_0^{\pi/2} \frac{\sqrt{\cos x}}{\sqrt{\cos x} + \sqrt{\sin x}} dx\] is
Given that \[\int\limits_0^\infty \frac{x^2}{\left( x^2 + a^2 \right)\left( x^2 + b^2 \right)\left( x^2 + c^2 \right)} dx = \frac{\pi}{2\left( a + b \right)\left( b + c \right)\left( c + a \right)},\] the value of \[\int\limits_0^\infty \frac{dx}{\left( x^2 + 4 \right)\left( x^2 + 9 \right)},\]
Evaluate the following integrals :-
\[\int_2^4 \frac{x^2 + x}{\sqrt{2x + 1}}dx\]
\[\int\limits_0^{\pi/4} e^x \sin x dx\]
\[\int\limits_0^{2\pi} \cos^7 x dx\]
\[\int\limits_0^\pi \frac{x}{a^2 \cos^2 x + b^2 \sin^2 x} dx\]
\[\int\limits_{- \pi/4}^{\pi/4} \left| \tan x \right| dx\]
\[\int\limits_0^{\pi/2} \frac{\cos^2 x}{\sin x + \cos x} dx\]
\[\int\limits_0^{\pi/2} \frac{x}{\sin^2 x + \cos^2 x} dx\]
\[\int\limits_0^2 \left( x^2 + 2 \right) dx\]
Using second fundamental theorem, evaluate the following:
`int_0^(pi/2) sqrt(1 + cos x) "d"x`
Evaluate the following:
`Γ (9/2)`
Evaluate the following:
`int_0^oo "e"^(-mx) x^6 "d"x`
Choose the correct alternative:
Using the factorial representation of the gamma function, which of the following is the solution for the gamma function Γ(n) when n = 8 is
Choose the correct alternative:
`Γ(3/2)`
`int (cos2x - cos 2theta)/(cosx - costheta) "d"x` is equal to ______.
`int "e"^x ((1 - x)/(1 + x^2))^2 "d"x` is equal to ______.
