मराठी

Π / 2 ∫ 0 Sin 2 X ( 1 + Cos X ) 2 D X

Advertisements
Advertisements

प्रश्न

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

बेरीज
Advertisements

उत्तर

\[\int_0^\frac{\pi}{2} \frac{\sin^2 x}{\left( 1 + \cos x \right)^2} d x\]

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

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

\[ = \int_0^\frac{\pi}{2} \frac{1 - \cos x}{1 + \cos x} d x\]

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

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

\[ = \int_0^\frac{\pi}{2} \frac{2}{1 + \cos x}dx - \int_0^\frac{\pi}{2} dx\]

\[ = \int_0^\frac{\pi}{2} \frac{2\left( 1 - \cos x \right)}{\left( 1 + \cos x \right)\left( 1 - \cos x \right)}dx - \int_0^\frac{\pi}{2} dx\]

\[ = 2 \int_0^\frac{\pi}{2} \frac{1 - \cos x}{\sin^2 x}dx - \left[ x \right]_0^\frac{\pi}{2} \]

\[ = 2 \int_0^\frac{\pi}{2} \left( \ cosec^2 x - \ cosec x\ cotx \right) dx - \left[ x \right]_0^\frac{\pi}{2} \]

\[ = 2 \left[ - cotx + \ cosec x \right]_0^\frac{\pi}{2} - \left[ x \right]_0^\frac{\pi}{2} \]

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

shaalaa.com
  या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
पाठ 19: Definite Integrals - Revision Exercise [पृष्ठ १२१]

APPEARS IN

आर.डी. शर्मा Mathematics Volume 1 and 2 [English] Class 12
पाठ 19 Definite Integrals
Revision Exercise | Q 11 | पृष्ठ १२१

संबंधित प्रश्‍न

\[\int\limits_0^{1/2} \frac{1}{\sqrt{1 - x^2}} dx\]

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

\[\int\limits_0^{\pi/6} \cos x \cos 2x\ dx\]

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

\[\int\limits_0^2 \frac{1}{\sqrt{3 + 2x - x^2}} dx\]

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

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

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

\[\int_0^1 x\log\left( 1 + 2x \right)dx\]

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

\[\int\limits_0^\pi 5 \left( 5 - 4 \cos \theta \right)^{1/4} \sin \theta\ d \theta\]

Evaluate the following integral:

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

\[\int_0^{2\pi} \cos^{- 1} \left( \cos x \right)dx\]

\[\int\limits_0^\pi x \sin^3 x\ dx\]

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

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

\[\int\limits_{- \pi/4}^{\pi/4} \sin^2 x\ dx\]

If f is an integrable function, show that

\[\int\limits_{- a}^a f\left( x^2 \right) dx = 2 \int\limits_0^a f\left( x^2 \right) dx\]


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

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

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


Write the coefficient abc of which the value of the integral

\[\int\limits_{- 3}^3 \left( a x^2 + bx + c \right) dx\] is independent.

\[\int\limits_0^1 \left\{ x \right\} dx,\] where {x} denotes the fractional part of x.  

 

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

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

 


If \[I_{10} = \int\limits_0^{\pi/2} x^{10} \sin x\ dx,\]  then the value of I10 + 90I8 is

 


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

 


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


\[\int\limits_1^5 \frac{x}{\sqrt{2x - 1}} dx\]


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


\[\int\limits_0^{\pi/2} \left| \sin x - \cos x \right| dx\]


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


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


Evaluate the following:

f(x) = `{{:("c"x",", 0 < x < 1),(0",",  "otherwise"):}` Find 'c" if `int_0^1 "f"(x)  "d"x` = 2


Evaluate the following using properties of definite integral:

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


Evaluate the following integrals as the limit of the sum:

`int_0^1 (x + 4)  "d"x`


Evaluate the following integrals as the limit of the sum:

`int_1^3 (2x + 3)  "d"x`


Choose the correct alternative:

`Γ(3/2)`


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


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×