Advertisements
Advertisements
प्रश्न
\[\int\limits_0^\pi \frac{1}{1 + \sin x} dx\] equals
पर्याय
0
1/2
2
3/2
Advertisements
उत्तर
2
\[\int_0^\pi \frac{1}{1 + \sin x} d x\]
\[ = \int_0^\pi \frac{1}{1 + \sin x} \times \frac{1 - \sin x}{1 - \sin x}dx\]
\[ = \int_0^\pi \frac{1 - \sin x}{1 - \sin^2 x}dx\]
\[ = \int_0^\pi \frac{1 - \sin x}{\cos^2 x}dx\]
\[ = \int_0^\pi \left( se c^2 x - \sec x \tan x \right) dx\]
\[ = \left[ \tan x - sec x \right]_0^\pi \]
\[ = 0 + 1 - 0 + 1\]
\[ = 2\]
APPEARS IN
संबंधित प्रश्न
\[\int\limits_1^4 f\left( x \right) dx, where f\left( x \right) = \begin{cases}7x + 3 & , & \text{if }1 \leq x \leq 3 \\ 8x & , & \text{if }3 \leq x \leq 4\end{cases}\]
Evaluate the following integral:
Evaluate :
If \[\left[ \cdot \right] and \left\{ \cdot \right\}\] denote respectively the greatest integer and fractional part functions respectively, evaluate the following integrals:
\[\int\limits_0^{\pi/2} \frac{1}{2 + \cos x} dx\] equals
The value of \[\int\limits_0^{\pi/2} \cos x\ e^{\sin x}\ dx\] is
If f (a + b − x) = f (x), then \[\int\limits_a^b\] x f (x) dx is equal to
\[\int\limits_0^4 x\sqrt{4 - x} dx\]
\[\int\limits_0^1 \cos^{- 1} x dx\]
\[\int\limits_0^1 \tan^{- 1} \left( \frac{2x}{1 - x^2} \right) dx\]
\[\int\limits_0^{\pi/2} \frac{\sin x}{\sqrt{1 + \cos x}} dx\]
\[\int\limits_0^\infty \frac{x}{\left( 1 + x \right)\left( 1 + x^2 \right)} dx\]
\[\int\limits_0^\pi \frac{x}{1 + \cos \alpha \sin x} dx\]
\[\int\limits_0^{\pi/2} \frac{\cos^2 x}{\sin x + \cos x} dx\]
\[\int\limits_0^1 \cot^{- 1} \left( 1 - x + x^2 \right) dx\]
Using second fundamental theorem, evaluate the following:
`int_0^1 x"e"^(x^2) "d"x`
Evaluate the following using properties of definite integral:
`int_0^(i/2) (sin^7x)/(sin^7x + cos^7x) "d"x`
Evaluate the following:
Γ(4)
Evaluate `int "dx"/sqrt((x - alpha)(beta - x)), beta > alpha`
Evaluate the following:
`int ((x^2 + 2))/(x + 1) "d"x`
