Advertisements
Advertisements
Question
Options
- \[\log\left( \frac{2}{3} \right)\]
- \[\log\left( \frac{3}{2} \right)\]
- \[\log\left( \frac{3}{4} \right)\]
- \[\log\left( \frac{4}{3} \right)\]
Advertisements
Solution
\[Let\, I = \int_0^\frac{\pi}{2} \frac{\cos x}{\left( 2 + \sin x \right)\left( 1 + \sin x \right)} d x\]
\[\text{Let} \sin x , \text{then} \cos x\ dx = dt\]
\[When\ x = 0, t = 0, x = \frac{\pi}{2}, t = 1\]
\[\text{Therefore the integral becomes}\]
\[I = \int_0^1 \frac{dt}{\left( 2 + t \right)\left( 1 + t \right)}\]
\[ = \int_0^1 \left[ \frac{- 1}{2 + t} + \frac{1}{1 + t} \right] dt\]
\[ = \left[ - \log\left( 2 + t \right) + \log\left( 1 + t \right) \right]_0^1 \]
\[ = \left[ \log\left( 1 + t \right) - \log\left( 2 + t \right) \right]_0^1 \]
\[ = \log2 - \log3 - \log1 + \log2\]
\[ = \log\frac{4}{3}\]
APPEARS IN
RELATED QUESTIONS
If f (x) is a continuous function defined on [0, 2a]. Then, prove that
If \[\int\limits_0^a 3 x^2 dx = 8,\] write the value of a.
If \[f\left( x \right) = \int_0^x t\sin tdt\], the write the value of \[f'\left( x \right)\]
\[\int\limits_0^{\pi/2} \left| \sin x - \cos x \right| dx\]
\[\int\limits_{- 1/2}^{1/2} \cos x \log\left( \frac{1 + x}{1 - x} \right) dx\]
\[\int\limits_0^\pi x \sin x \cos^4 x dx\]
\[\int\limits_{- \pi}^\pi x^{10} \sin^7 x dx\]
\[\int\limits_1^3 \left( x^2 + 3x \right) dx\]
Prove that `int_a^b ƒ ("x") d"x" = int_a^bƒ(a + b - "x") d"x" and "hence evaluate" int_(π/6)^(π/3) (d"x")/(1+sqrt(tan "x")`
Evaluate the following:
`int_0^2 "f"(x) "d"x` where f(x) = `{{:(3 - 2x - x^2",", x ≤ 1),(x^2 + 2x - 3",", 1 < x ≤ 2):}`
Evaluate the following integrals as the limit of the sum:
`int_1^3 x "d"x`
Choose the correct alternative:
If f(x) is a continuous function and a < c < b, then `int_"a"^"c" f(x) "d"x + int_"c"^"b" f(x) "d"x` is
Choose the correct alternative:
`Γ(3/2)`
Evaluate `int sqrt((1 + x)/(1 - x)) "d"x`, x ≠1
Evaluate `int "dx"/sqrt((x - alpha)(beta - x)), beta > alpha`
Evaluate `int (x^2 + x)/(x^4 - 9) "d"x`
`int "e"^x ((1 - x)/(1 + x^2))^2 "d"x` is equal to ______.
