मराठी

Π ∫ 0 1 1 + Sin X D X Equals(A) 0 (B) 1/2 (C) 2 (D) 3/2

Advertisements
Advertisements

प्रश्न

\[\int\limits_0^\pi \frac{1}{1 + \sin x} dx\] equals

पर्याय

  • 0

  • 1/2

  • 2

  • 3/2

MCQ
Advertisements

उत्तर

2

 

\[\int_0^\pi \frac{1}{1 + \sin x} d x\]
\[ = \int_0^\pi \frac{1}{1 + \sin x} \times \frac{1 - \sin x}{1 - \sin x}dx\]
\[ = \int_0^\pi \frac{1 - \sin x}{1 - \sin^2 x}dx\]
\[ = \int_0^\pi \frac{1 - \sin x}{\cos^2 x}dx\]
\[ = \int_0^\pi \left( se c^2 x - \sec x \tan x \right) dx\]
\[ = \left[ \tan x - sec x \right]_0^\pi \]
\[ = 0 + 1 - 0 + 1\]
\[ = 2\]

shaalaa.com
  या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
पाठ 19: Definite Integrals - MCQ [पृष्ठ ११७]

APPEARS IN

आर.डी. शर्मा Mathematics Volume 1 and 2 [English] Class 12
पाठ 19 Definite Integrals
MCQ | Q 2 | पृष्ठ ११७

संबंधित प्रश्‍न

\[\int\limits_2^3 \frac{x}{x^2 + 1} dx\]

\[\int\limits_{- \pi/4}^{\pi/4} \frac{1}{1 + \sin x} dx\]

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

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

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

\[\int\limits_1^e \frac{\log x}{x} dx\]

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

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

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

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

\[\int_0^\frac{\pi}{2} \frac{\cos^2 x}{1 + 3 \sin^2 x}dx\]

Evaluate the following integral:

\[\int\limits_{- 3}^3 \left| x + 1 \right| dx\]

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

\[\int\limits_0^{\pi/2} \frac{\sin^{3/2} x}{\sin^{3/2} x + \cos^{3/2} x} dx\]

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

 


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

\[\int\limits_{- 1}^1 \left( x + 3 \right) dx\]

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

Evaluate each of the following integral:

\[\int_0^\frac{\pi}{4} \sin2xdx\]

\[\int\limits_0^2 x\left[ x \right] dx .\]

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


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

\[\int\limits_1^\sqrt{3} \frac{1}{1 + x^2} dx\]  is equal to ______.

The value of \[\int\limits_{- \pi}^\pi \sin^3 x \cos^2 x\ dx\] is 

 


The value of \[\int\limits_0^{\pi/2} \log\left( \frac{4 + 3 \sin x}{4 + 3 \cos x} \right) dx\] is 

 


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


Evaluate : \[\int\frac{dx}{\sin^2 x \cos^2 x}\] .


\[\int\limits_0^4 x\sqrt{4 - x} dx\]


\[\int\limits_1^2 x\sqrt{3x - 2} dx\]


\[\int\limits_0^{\pi/3} \frac{\cos x}{3 + 4 \sin x} dx\]


\[\int\limits_0^{\pi/4} \tan^4 x dx\]


\[\int\limits_{- \pi/2}^{\pi/2} \sin^9 x dx\]


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


\[\int\limits_0^\pi \frac{dx}{6 - \cos x}dx\]


Using second fundamental theorem, evaluate the following:

`int_0^(1/4) sqrt(1 - 4)  "d"x`


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


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×