Advertisements
Advertisements
प्रश्न
The value of \[\int\limits_0^\pi \frac{1}{5 + 3 \cos x} dx\] is
पर्याय
π/4
π/8
π/2
0
Advertisements
उत्तर
π/4
\[\int_0^\pi \frac{1}{5 + 3 \cos x} d x\]
\[ = \int_0^\pi \frac{1}{5 + 3 \frac{1 - \tan^2 \frac{x}{2}}{1 + \tan^2 \frac{x}{2}}} d x\]
\[ = \int_0^\pi \frac{1 + \tan^2 \frac{x}{2}}{5 + 5 \tan^2 \frac{x}{2} + 3 - 3 \tan^2 \frac{x}{2}}dx\]
\[ = \int_0^\pi \frac{se c^2 \frac{x}{2}}{8 + 2 \tan^2 \frac{x}{2}}dx\]
\[Let\ \tan\frac{x}{2} = t, \text{then }\sec^2 \frac{x}{2} dx = 2dt\]
\[When\ x = 0, t = 0, x = \pi, t = \infty \]
\[\text{Therefore the integral becomes}\]
\[\frac{1}{2} \int_0^\infty \frac{dt}{4 + t^2}\]
\[ = \frac{1}{2} \left[ \tan^{- 1} \frac{t}{2} \right]_0^\infty \]
\[ = \frac{1}{2}\left( \frac{\pi}{2} - 0 \right) = \frac{\pi}{4}\]
APPEARS IN
संबंधित प्रश्न
If f(2a − x) = −f(x), prove that
\[\int\limits_0^1 \left\{ x \right\} dx,\] where {x} denotes the fractional part of x.
The value of \[\int\limits_0^{\pi/2} \log\left( \frac{4 + 3 \sin x}{4 + 3 \cos x} \right) dx\] is
\[\int\limits_1^2 \frac{x + 3}{x\left( x + 2 \right)} dx\]
\[\int\limits_0^1 \left| 2x - 1 \right| dx\]
\[\int\limits_0^\pi \cos 2x \log \sin x dx\]
\[\int\limits_0^1 \cot^{- 1} \left( 1 - x + x^2 \right) dx\]
\[\int\limits_0^{\pi/2} \frac{dx}{4 \cos x + 2 \sin x}dx\]
\[\int\limits_0^4 x dx\]
\[\int\limits_1^3 \left( x^2 + 3x \right) dx\]
Evaluate the following using properties of definite integral:
`int_0^1 x/((1 - x)^(3/4)) "d"x`
Evaluate the following:
`int_0^oo "e"^(-mx) x^6 "d"x`
Evaluate the following:
`int_0^oo "e"^(-4x) x^4 "d"x`
Evaluate the following:
`int_0^oo "e"^(- x/2) x^5 "d"x`
Evaluate the following integrals as the limit of the sum:
`int_1^3 x "d"x`
Choose the correct alternative:
`int_0^oo "e"^(-2x) "d"x` is
Choose the correct alternative:
Γ(n) is
Evaluate `int (x^2 + x)/(x^4 - 9) "d"x`
