हिंदी

The Value of π / 2 ∫ − π / 2 ( X 3 + X Cos X + Tan 5 X + 1 ) D X , Is,0,2,π,1

Advertisements
Advertisements

प्रश्न

The value of \[\int\limits_{- \pi/2}^{\pi/2} \left( x^3 + x \cos x + \tan^5 x + 1 \right) dx, \] is 

विकल्प

  •  0

  • 2

  • π

  • 1

MCQ
Advertisements

उत्तर

π

\[\int_{- \frac{\pi}{2}}^\frac{\pi}{2} \left( x^3 + x\cos x + \tan^5 x + 1 \right) d x\]

\[ = \left[ \frac{x^4}{4} \right]_{- \frac{\pi}{2}}^\frac{\pi}{2} + \left[ x \sin x \right]_{- \frac{\pi}{2}}^\frac{\pi}{2} - \int_{- \frac{\pi}{2}}^\frac{\pi}{2} \sin x dx + \int_{- \frac{\pi}{2}}^\frac{\pi}{2} \tan^3 x \left( se c^2 x - 1 \right)dx + \left[ x \right]_{- \frac{\pi}{2}}^\frac{\pi}{2} \]

\[ = \frac{\pi^4}{64} - \frac{\pi^4}{64} + \frac{\pi}{2} - \frac{\pi}{2} - \left[ - \cos x \right]_{- \frac{\pi}{2}}^\frac{\pi}{2} + \int_{- \frac{\pi}{2}}^\frac{\pi}{2} \tan^3 x se c^2 x dx - \int_{- \frac{\pi}{2}}^\frac{\pi}{2} \tan^3 x dx + \frac{\pi}{2} + \frac{\pi}{2}\]

\[ = \pi + 0 + \left[ \frac{\tan^4 x}{4} \right]_{- \frac{\pi}{2}}^\frac{\pi}{2} - \int_{- \frac{\pi}{2}}^\frac{\pi}{2} tanx \sec^2 x dx - \int_{- \frac{\pi}{2}}^\frac{\pi}{2} tan x dx\]

\[ = \pi - \left[ \frac{\tan^2 x}{2} \right]_{- \frac{\pi}{2}}^\frac{\pi}{2} - \left[ - \log\left( \cos x \right) \right]_{- \frac{\pi}{2}}^\frac{\pi}{2} \]

\[ = \pi\]

shaalaa.com
  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 19: Definite Integrals - MCQ [पृष्ठ १२०]

APPEARS IN

आर.डी. शर्मा Mathematics Volume 1 and 2 [English] Class 12
अध्याय 19 Definite Integrals
MCQ | Q 42 | पृष्ठ १२०

संबंधित प्रश्न

\[\int\limits_4^9 \frac{1}{\sqrt{x}} dx\]

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

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

\[\int\limits_{\pi/2}^\pi e^x \left( \frac{1 - \sin x}{1 - \cos x} \right) dx\]

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

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

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

\[\int\limits_0^{\pi/2} \sqrt{\sin \phi} \cos^5 \phi\ d\phi\]

 


\[\int\limits_0^{\pi/2} \frac{1}{5 + 4 \sin x} dx\]

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

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

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

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

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

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

Evaluate the following integral:

\[\int_{- a}^a \log\left( \frac{a - \sin\theta}{a + \sin\theta} \right)d\theta\]

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

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

\[\int\limits_0^{\pi/2} \log \tan x\ dx .\]

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

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

\[\int\limits_0^\infty \frac{1}{1 + e^x} dx\]  equals


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


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


\[\int\limits_0^{\pi/2} \frac{1}{1 + \cot^7 x} dx\]


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


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


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


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


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


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


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


Using second fundamental theorem, evaluate the following:

`int_0^1 "e"^(2x)  "d"x`


Using second fundamental theorem, evaluate the following:

`int_1^2 (x - 1)/x^2  "d"x`


Evaluate the following:

`int_0^oo "e"^(-mx) x^6 "d"x`


Choose the correct alternative:

`Γ(3/2)`


`int (cos2x - cos 2theta)/(cosx - costheta) "d"x` is equal to ______.


`int x^3/(x + 1)` is equal to ______.


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×