मराठी
महाराष्ट्र राज्य शिक्षण मंडळएचएससी विज्ञान (सामान्य) इयत्ता १२ वी

Evaluate the following : ∫2+x2-x.dx

Advertisements
Advertisements

प्रश्न

Evaluate the following : `int sqrt((2 + x)/(2 - x)).dx`

बेरीज
Advertisements

उत्तर

Let I = `int sqrt((2 + x)/(2 - x)).dx`

= `int sqrt((2 + x)/(2 - x) xx (2 + x)/(2 + x)).dx`

= `int (2 + x)/sqrt(4 - x^2).dx`

= `int (2)/sqrt(4 - x^2).dx + int x/sqrt(4 - x^2).dx`

= `2 int (1)/sqrt(2^2 - x^2).dx + (1)/(2) int (2x)/sqrt(4 - x^2).dx`

= I1 + I2                        ...(Let)

I1 = `2 int (1)/sqrt(2^2 - x^2).dx`

= `2 sin^-1 (x/2) + c_1`

In I2, put 4 – x2 = t
∴ – 2x dx =  dt
∴  2x dx = – dt

I2 = `-(1)/(2) int t^(-1/2) dt`

= `-(1)/(2).t^(1/2)/((1/2)) + c_2`

= `- sqrt(4 - x^2) + c_2`

I = `2 sin^-1 (x/2) - sqrt(4 - x^2) + c`.

shaalaa.com
  या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
पाठ 3: Indefinite Integration - Exercise 3.2 (B) [पृष्ठ १२३]

APPEARS IN

बालभारती Mathematics and Statistics 2 (Arts and Science) [English] Standard 12 Maharashtra State Board
पाठ 3 Indefinite Integration
Exercise 3.2 (B) | Q 1.08 | पृष्ठ १२३

व्हिडिओ ट्यूटोरियलVIEW ALL [2]

संबंधित प्रश्‍न

Evaluate :   `∫1/(cos^4x+sin^4x)dx`


Integrate the functions:

(4x + 2) `sqrt(x^2 + x +1)`


Integrate the functions:

`(e^(2x) - 1)/(e^(2x) + 1)`


Integrate the functions:

`sqrt(sin 2x) cos 2x`


Integrate the functions:

`sqrt(tanx)/(sinxcos x)`


Evaluate `int 1/(3+ 2 sinx + cosx) dx`


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

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

Write a value of

\[\int x^2 \sin x^3 \text{ dx }\]

Write a value of

\[\int\frac{\cos x}{3 + 2 \sin x}\text{  dx}\]

Write a value of\[\int\left( e^{x \log_e \text{  a}} + e^{a \log_e x} \right) dx\] .


Write a value of\[\int\frac{\cos x}{\sin x \log \sin x} dx\]

 


Write a value of\[\int e^x \left( \frac{1}{x} - \frac{1}{x^2} \right) dx\] .


Evaluate:  \[\int\frac{x^3 - 1}{x^2} \text{ dx}\]


\[\text{ If } \int\left( \frac{x - 1}{x^2} \right) e^x dx = f\left( x \right) e^x + C, \text{ then  write  the value of  f}\left( x \right) .\]

 Show that : `int _0^(pi/4) "log" (1+"tan""x")"dx" = pi /8 "log"2`


Integrate the following functions w.r.t. x : `e^x.log (sin e^x)/tan(e^x)`


Integrate the following functions w.r.t. x : `(cos3x - cos4x)/(sin3x + sin4x)`


Integrate the following functions w.r.t. x : `cosx/sin(x - a)`


Integrate the following functions w.r.t. x : `3^(cos^2x) sin 2x`


Evaluate the following:

`int (1)/(25 - 9x^2)*dx`


Evaluate the following : `int  (1)/(x^2 + 8x + 12).dx`


Evaluate the following : `int (1)/(cos2x + 3sin^2x).dx`


Evaluate the following integrals :  `int (3x + 4)/sqrt(2x^2 + 2x + 1).dx`


Evaluate the following integrals : `int sqrt((9 - x)/x).dx`


Choose the correct options from the given alternatives :

`int sqrt(cotx)/(sinx*cosx)*dx` =


Choose the correct options from the given alternatives :

`int (e^(2x) + e^-2x)/e^x*dx` =


Evaluate the following.

`int 1/(sqrt("x"^2 -8"x" - 20))` dx


Choose the correct alternative from the following.

`int "x"^2 (3)^("x"^3) "dx"` =


Evaluate: `int sqrt("x"^2 + 2"x" + 5)` dx


`int 1/sqrt((x - 3)(x + 2))` dx = ______.


`int 1/(cos x - sin x)` dx = _______________


`int e^x/x [x (log x)^2 + 2 log x]` dx = ______________


`int 2/(sqrtx - sqrt(x + 3))` dx = ________________


`int (2(cos^2 x - sin^2 x))/(cos^2 x + sin^2 x)` dx = ______________


`int ("e"^(3x))/("e"^(3x) + 1)  "d"x`


Evaluate  `int"e"^x (1/x - 1/x^2)  "d"x`


If `int x^3"e"^(x^2) "d"x = "e"^(x^2)/2 "f"(x) + "c"`, then f(x) = ______.


`int (sin  (5x)/2)/(sin  x/2)dx` is equal to ______. (where C is a constant of integration).


`int(log(logx) + 1/(logx)^2)dx` = ______.


`int sqrt(x^2 - a^2)/x dx` = ______.


Evaluate `int(1 + x + x^2/(2!) )dx`


if `f(x) = 4x^3 - 3x^2 + 2x +k, f (0) = - 1 and f (1) = 4, "find " f(x)`


If f'(x) = 4x3- 3x2 + 2x + k, f(0) = 1 and f(1) = 4, find f(x).


`int (cos4x)/(sin2x + cos2x)dx` = ______.


Evaluate the following.

`intxsqrt(1+x^2)dx`


Evaluate the following.

`int1/(x^2+4x-5)dx`


What is \[\int\frac{\sin x}{\sin(x+a)}\,dx\]?


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×