Advertisements
Advertisements
प्रश्न
\[\int\limits_0^{\pi/2} \frac{\sin x}{\sqrt{1 + \cos x}} dx\]
Advertisements
उत्तर
\[\int_0^\frac{\pi}{2} \frac{\sin x}{\sqrt{1 + \cos x}} d x\]
\[Let 1 + \cos x = t, then - \sin x dx = dt\]
\[When, x \to 0, t \to 2 and x \to \frac{\pi}{2}, t \to 1\]
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
संबंधित प्रश्न
If f is an integrable function, show that
\[\int\limits_{- a}^a f\left( x^2 \right) dx = 2 \int\limits_0^a f\left( x^2 \right) dx\]
Evaluate each of the following integral:
\[\int\limits_0^1 \left\{ x \right\} dx,\] where {x} denotes the fractional part of x.
The value of the integral \[\int\limits_0^\infty \frac{x}{\left( 1 + x \right)\left( 1 + x^2 \right)} dx\]
Evaluate : \[\int\limits_0^\pi/4 \frac{\sin x + \cos x}{16 + 9 \sin 2x}dx\] .
Evaluate : \[\int\limits_0^\pi \frac{x}{1 + \sin \alpha \sin x}dx\] .
Evaluate : \[\int\frac{dx}{\sin^2 x \cos^2 x}\] .
\[\int\limits_0^1 \frac{1 - x}{1 + x} dx\]
\[\int\limits_1^2 \frac{1}{x^2} e^{- 1/x} dx\]
\[\int\limits_{\pi/3}^{\pi/2} \frac{\sqrt{1 + \cos x}}{\left( 1 - \cos x \right)^{5/2}} dx\]
\[\int\limits_{- \pi/4}^{\pi/4} \left| \tan x \right| dx\]
\[\int\limits_2^3 \frac{\sqrt{x}}{\sqrt{5 - x} + \sqrt{x}} dx\]
\[\int\limits_0^{\pi/2} \frac{x}{\sin^2 x + \cos^2 x} dx\]
\[\int\limits_0^4 x dx\]
\[\int\limits_{- 1}^1 e^{2x} dx\]
Using second fundamental theorem, evaluate the following:
`int_0^1 "e"^(2x) "d"x`
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:
Γ(4)
Evaluate the following integrals as the limit of the sum:
`int_1^3 (2x + 3) "d"x`
Choose the correct alternative:
The value of `int_(- pi/2)^(pi/2) cos x "d"x` is
Choose the correct alternative:
If n > 0, then Γ(n) is
`int x^9/(4x^2 + 1)^6 "d"x` is equal to ______.
