मराठी

Π / 2 ∫ 0 1 1 + Tan 3 X D X - Mathematics

Advertisements
Advertisements

प्रश्न

\[\int\limits_0^{\pi/2} \frac{1}{1 + \tan^3 x} dx\]

बेरीज
Advertisements

उत्तर

\[Let, I = \int_0^\frac{\pi}{2} \frac{1}{1 + \tan^3 x} d x ..............(1)\]

\[ = \int_0^\frac{\pi}{2} \frac{1}{1 + \tan^3 \left( \frac{\pi}{2} - x \right)} d x\]

\[ = \int_0^\frac{\pi}{2} \frac{1}{1 + co t^3 x} d x ................(2)\]

Adding (1) and (2)

\[2I = \int_0^\frac{\pi}{2} \left[ \frac{1}{1 + \tan^3 x} + \frac{1}{1 + co t^3 x} \right] d x\]

\[ = \int_0^\frac{\pi}{2} \frac{2 + \tan^3 x + co t^3 x}{\left( 1 + \tan^3 x \right)\left( 1 + co t^3 x \right)}dx\]

\[ = \int_0^\frac{\pi}{2} \frac{2 + \tan^3 x + co t^3 x}{2 + \tan^3 x + co t^3 x}dx\]

\[ = \int_0^\frac{\pi}{2} dx \]

\[ = \left( x \right)_0^\frac{\pi}{2} \]

\[ = \frac{\pi}{2}\]

\[Hence, I = \frac{\pi}{4}\]

shaalaa.com
Definite Integrals
  या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
पाठ 20: Definite Integrals - Revision Exercise [पृष्ठ १२२]

APPEARS IN

आरडी शर्मा Mathematics [English] Class 12
पाठ 20 Definite Integrals
Revision Exercise | Q 40 | पृष्ठ १२२

संबंधित प्रश्‍न

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

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

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

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

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

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

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

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

Evaluate each of the following integral:

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

 


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

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

Evaluate the following integral:

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

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

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

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

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

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

 


\[\int\limits_0^\infty e^{- x} dx .\]

Evaluate each of the following integral:

\[\int_0^\frac{\pi}{4} \tan\ xdx\]

 


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


\[\int\limits_0^1 e^\left\{ x \right\} dx .\]

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

 


Evaluate : \[\int e^{2x} \cdot \sin \left( 3x + 1 \right) dx\] .


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


\[\int\limits_0^{\pi/4} \sin 2x \sin 3x dx\]


\[\int\limits_0^1 \left| \sin 2\pi x \right| dx\]


\[\int\limits_0^{2\pi} \cos^7 x dx\]


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


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


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


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


Find : `∫_a^b logx/x` dx


Evaluate the following using properties of definite integral:

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


Evaluate the following:

`int_0^oo "e"^(-4x) x^4  "d"x`


Evaluate `int (3"a"x)/("b"^2 + "c"^2x^2) "d"x`


If `int (3"e"^x - 5"e"^-x)/(4"e"6x + 5"e"^-x)"d"x` = ax + b log |4ex + 5e –x| + C, then ______.


Evaluate the following:

`int ((x^2 + 2))/(x + 1) "d"x`


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


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×