मराठी

Π ∫ 0 / 2 Cos X ( 2 + Sin X ) ( 1 + Sin X ) D X Equals,Log ( 2 3 ) ,Log ( 3 2 ),Log ( 3 4 ),Log ( 4 3 ) - Mathematics

Advertisements
Advertisements

प्रश्न

\[\int\limits_0^{\pi/2} \frac{\cos x}{\left( 2 + \sin x \right)\left( 1 + \sin x \right)} dx\] equals

पर्याय

  • \[\log\left( \frac{2}{3} \right)\]
  • \[\log\left( \frac{3}{2} \right)\]
  • \[\log\left( \frac{3}{4} \right)\]
  • \[\log\left( \frac{4}{3} \right)\]
MCQ

उत्तर

\[\log\left( \frac{4}{3} \right)\]

\[Let\, I = \int_0^\frac{\pi}{2} \frac{\cos x}{\left( 2 + \sin x \right)\left( 1 + \sin x \right)} d x\]
\[\text{Let} \sin x , \text{then} \cos x\ dx = dt\]
\[When\ x = 0, t = 0, x = \frac{\pi}{2}, t = 1\]
\[\text{Therefore the integral becomes}\]
\[I = \int_0^1 \frac{dt}{\left( 2 + t \right)\left( 1 + t \right)}\]
\[ = \int_0^1 \left[ \frac{- 1}{2 + t} + \frac{1}{1 + t} \right] dt\]
\[ = \left[ - \log\left( 2 + t \right) + \log\left( 1 + t \right) \right]_0^1 \]
\[ = \left[ \log\left( 1 + t \right) - \log\left( 2 + t \right) \right]_0^1 \]
\[ = \log2 - \log3 - \log1 + \log2\]
\[ = \log\frac{4}{3}\]

shaalaa.com
Definite Integrals
  या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
पाठ 20: Definite Integrals - MCQ [पृष्ठ ११७]

APPEARS IN

आरडी शर्मा Mathematics [English] Class 12
पाठ 20 Definite Integrals
MCQ | Q 8 | पृष्ठ ११७

संबंधित प्रश्‍न

\[\int\limits_0^{1/2} \frac{1}{\sqrt{1 - x^2}} dx\]

\[\int\limits_{- 1}^1 \frac{1}{1 + x^2} dx\]

\[\int\limits_0^{\pi/2} x \cos\ x\ dx\]

\[\int\limits_e^{e^2} \left\{ \frac{1}{\log x} - \frac{1}{\left( \log x \right)^2} \right\} dx\]

\[\int\limits_0^2 \frac{1}{4 + x - x^2} dx\]

\[\int\limits_1^3 \frac{\cos \left( \log x \right)}{x} dx\]

\[\int\limits_0^a \sqrt{a^2 - x^2} dx\]

\[\int\limits_0^{\pi/2} \frac{\sin \theta}{\sqrt{1 + \cos \theta}} d\theta\]

\[\int\limits_0^1 \tan^{- 1} \left( \frac{2x}{1 - x^2} \right) dx\]

\[\int\limits_0^1 \sqrt{\frac{1 - x}{1 + x}} dx\]

\[\int\limits_0^{\pi/4} \sin^3 2t \cos 2t\ dt\]

\[\int\limits_0^{\pi/6} \cos^{- 3} 2 \theta \sin 2\ \theta\ d\ \theta\]

\[\int\limits_0^{( \pi )^{2/3}} \sqrt{x} \cos^2 x^{3/2} dx\]


\[\int\limits_0^\pi \sin^3 x\left( 1 + 2 \cos x \right) \left( 1 + \cos x \right)^2 dx\]

\[\int\limits_0^a \sin^{- 1} \sqrt{\frac{x}{a + x}} dx\]

\[\int\limits_{- a}^a \sqrt{\frac{a - x}{a + x}} dx\]

\[\int\limits_{- 1}^1 \log\left( \frac{2 - x}{2 + x} \right) dx\]

\[\int\limits_{- \pi/2}^{\pi/2} \log\left( \frac{2 - \sin x}{2 + \sin x} \right) dx\]

\[\int\limits_1^2 x^2 dx\]

\[\int\limits_2^3 \left( 2 x^2 + 1 \right) dx\]

\[\int\limits_0^2 \left( x^2 + 2x + 1 \right) dx\]

\[\int\limits_0^1 \frac{1}{x^2 + 1} dx\]

\[\int\limits_{- 2}^1 \frac{\left| x \right|}{x} dx .\]

\[\int\limits_2^3 \frac{1}{x}dx\]

Solve each of the following integral:

\[\int_2^4 \frac{x}{x^2 + 1}dx\]

If \[\int\limits_0^1 \left( 3 x^2 + 2x + k \right) dx = 0,\] find the value of k.

 


If \[f\left( x \right) = \int_0^x t\sin tdt\], the write the value of \[f'\left( x \right)\]


If \[\int_0^a \frac{1}{4 + x^2}dx = \frac{\pi}{8}\] , find the value of a.


The value of \[\int\limits_0^{2\pi} \sqrt{1 + \sin\frac{x}{2}}dx\] is 


The value of \[\int\limits_0^\pi \frac{1}{5 + 3 \cos x} dx\] is

 


\[\int\limits_1^5 \frac{x}{\sqrt{2x - 1}} dx\]


\[\int\limits_0^1 x \left( \tan^{- 1} x \right)^2 dx\]


\[\int\limits_0^{15} \left[ x^2 \right] dx\]


\[\int\limits_2^3 \frac{\sqrt{x}}{\sqrt{5 - x} + \sqrt{x}} dx\]


\[\int\limits_0^2 \left( 2 x^2 + 3 \right) dx\]


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:

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


Integrate `((2"a")/sqrt(x) - "b"/x^2 + 3"c"root(3)(x^2))` w.r.t. x


Evaluate `int (x^2 + x)/(x^4 - 9) "d"x`


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×