हिंदी

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

विकल्प

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

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
  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 20: Definite Integrals - MCQ [पृष्ठ ११७]

APPEARS IN

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

संबंधित प्रश्न

\[\int\limits_0^{\pi/4} \left( \sqrt{\tan}x + \sqrt{\cot}x \right) dx\]

\[\int\limits_0^{\pi/2} \frac{1}{a^2 \sin^2 x + b^2 \cos^2 x} dx\]

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

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

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

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

\[\int_{- 1}^2 \left( \left| x + 1 \right| + \left| x \right| + \left| x - 1 \right| \right)dx\]

 


Evaluate each of the following integral:

\[\int_0^{2\pi} \log\left( \sec x + \tan x \right)dx\]

 


\[\int\limits_0^5 \frac{\sqrt[4]{x + 4}}{\sqrt[4]{x + 4} + \sqrt[4]{9 - x}} dx\]

If  \[f\left( a + b - x \right) = f\left( x \right)\] , then prove that \[\int_a^b xf\left( x \right)dx = \frac{a + b}{2} \int_a^b f\left( x \right)dx\]

 


\[\int\limits_0^{\pi/2} \frac{1}{1 + \tan x}\]

 


\[\int\limits_0^{\pi/2} \frac{1}{1 + \sqrt{\tan x}} dx\]

\[\int\limits_0^{\pi/2} \frac{x \sin x \cos x}{\sin^4 x + \cos^4 x} dx\]

\[\int\limits_0^\pi \log\left( 1 - \cos x \right) dx\]

Evaluate the following integral:

\[\int_{- 1}^1 \left| xcos\pi x \right|dx\]

 


If f is an integrable function, show that

\[\int\limits_{- a}^a x f\left( x^2 \right) dx = 0\]

 


Prove that:

\[\int_0^\pi xf\left( \sin x \right)dx = \frac{\pi}{2} \int_0^\pi f\left( \sin x \right)dx\]

\[\int\limits_1^3 \left( 2x + 3 \right) dx\]

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

\[\int\limits_a^b x\ dx\]

\[\int\limits_1^3 \left( 2 x^2 + 5x \right) dx\]

\[\int\limits_{- \pi/2}^{\pi/2} \sin^2 x\ dx .\]

Solve each of the following integral:

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

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


Evaluate : 

\[\int\limits_2^3 3^x dx .\]

The value of \[\int\limits_0^\pi \frac{x \tan x}{\sec x + \cos x} dx\] is __________ .


\[\int\limits_0^3 \frac{3x + 1}{x^2 + 9} dx =\]

Evaluate : \[\int\limits_0^\pi/4 \frac{\sin x + \cos x}{16 + 9 \sin 2x}dx\] .


Evaluate : \[\int\limits_0^\pi \frac{x}{1 + \sin \alpha \sin x}dx\] .


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


\[\int\limits_{\pi/6}^{\pi/2} \frac{\ cosec x \cot x}{1 + {cosec}^2 x} dx\]


\[\int\limits_0^4 x dx\]


Find : `∫_a^b logx/x` dx


Evaluate the following using properties of definite integral:

`int_(- pi/2)^(pi/2) sin^2theta  "d"theta`


Evaluate the following:

`Γ (9/2)`


Evaluate the following integrals as the limit of the sum:

`int_1^3 (2x + 3)  "d"x`


Choose the correct alternative:

`int_(-1)^1 x^3 "e"^(x^4)  "d"x` is


Evaluate `int "dx"/sqrt((x - alpha)(beta - x)), beta > alpha`


Verify the following:

`int (2x + 3)/(x^2 + 3x) "d"x = log|x^2 + 3x| + "C"`


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×