हिंदी

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

Advertisements
Advertisements

प्रश्न

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

विकल्प

  • 0

  • 1/2

  • 2

  • 3/2

MCQ

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
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 2 | पृष्ठ ११७

संबंधित प्रश्न

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

\[\int\limits_{\pi/3}^{\pi/4} \left( \tan x + \cot x \right)^2 dx\]

\[\int\limits_{- 1}^1 \frac{1}{x^2 + 2x + 5} dx\]

\[\int\limits_1^4 \frac{x^2 + x}{\sqrt{2x + 1}} dx\]

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

\[\int_0^\frac{\pi}{4} \left( \tan x + \cot x \right)^{- 2} dx\]

\[\int_0^1 \frac{1}{1 + 2x + 2 x^2 + 2 x^3 + x^4}dx\]

\[\int\limits_1^2 \frac{1}{x \left( 1 + \log x \right)^2} dx\]

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

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

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

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

 


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

\[\int\limits_1^4 \left( x^2 - x \right) dx\]

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

\[\int\limits_0^4 \frac{1}{\sqrt{16 - x^2}} dx .\]

\[\int\limits_0^{\pi/2} \frac{\sin^n x}{\sin^n x + \cos^n x} dx, n \in N .\]

\[\int\limits_0^\pi \cos^5 x\ dx .\]

\[\int\limits_0^2 \sqrt{4 - x^2} dx\]

Evaluate each of the following  integral:

\[\int_0^1 x e^{x^2} dx\]

 


Solve each of the following integral:

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

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

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


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


If \[\int\limits_0^a \frac{1}{1 + 4 x^2} dx = \frac{\pi}{8},\] then a equals

 


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

 


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


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


Evaluate the following integrals :-

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


\[\int\limits_0^\pi \cos 2x \log \sin x dx\]


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


\[\int\limits_1^4 \left( x^2 + x \right) dx\]


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


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")`


Using second fundamental theorem, evaluate the following:

`int_(-1)^1 (2x + 3)/(x^2 + 3x + 7)  "d"x`


Evaluate the following:

`int_0^oo "e"^(-4x) x^4  "d"x`


Choose the correct alternative:

`int_0^1 (2x + 1)  "d"x` is


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


Verify the following:

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


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×