मराठी

Evaluate the integral by using substitution. ∫02dxx+4-x2

Advertisements
Advertisements

प्रश्न

Evaluate the integral by using substitution.

`int_0^2 dx/(x + 4 - x^2)`

बेरीज
Advertisements

उत्तर

Let `I = int_0^2  dx/(x + 4 - x^2)`

`= int_0^2 dx/(4 - (x^2 - x))`

`= int_0^2 dx/(4 + 1/4 - (x - 1/2)^2)`

`= int_0^2 dx/((sqrt17/2)^2 - (x - 1/2)^2)`

`= 1/(2 xx sqrt17/2) [log  (sqrt17/2 + (x - 1/2))/(sqrt17/2 - (x - 1/2)}]_0^2`

`= 1/sqrt17 [log  (sqrt17 + 2x - 1)/(sqrt17 - 2x  + 1)]_0^2`

`= 1/sqrt17 [log  (sqrt17 + 3)/(sqrt17 - 3) - log  (sqrt17 - 1)/(sqrt17 + 1)]`

`= 1/sqrt17  log [(sqrt17 + 3)/(sqrt17 - 3) xx (sqrt17 + 1)/(sqrt17 - 1)]`

`= 1/sqrt17 log [(17 +3 + 3sqrt17 + sqrt17)/(17 + 3 - 3sqrt17 - sqrt17)]`

`= 1/sqrt17  log ((20 + 4sqrt17)/(20 - 4sqrt17))`

`= 1/sqrt17  log ((5 + sqrt17)/(5 - sqrt17))`

`= 1/sqrt17  log ((5 + sqrt17)/(5 - sqrt17) xx (5 + sqrt17)/(5 + sqrt17))`

`= 1/sqrt17  log [(25 + 17 + 10sqrt17)/(25 - 17)]`

`= 1/sqrt17  log  [(41 + 10 sqrt17)/8]`

`= 1/sqrt17  log [(21 + 5 sqrt17)/4]`

shaalaa.com
  या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
पाठ 7: Integrals - Exercise 7.10 [पृष्ठ ३४०]

APPEARS IN

एनसीईआरटी Mathematics Part 1 and 2 [English] Class 12
पाठ 7 Integrals
Exercise 7.10 | Q 6 | पृष्ठ ३४०

व्हिडिओ ट्यूटोरियलVIEW ALL [1]

संबंधित प्रश्‍न

Evaluate: `int1/(xlogxlog(logx))dx`


Evaluate : `int_0^4(|x|+|x-2|+|x-4|)dx`


Evaluate: `intsinsqrtx/sqrtxdx`

 


Evaluate the integral by using substitution.

`int_0^1 x/(x^2 +1)`dx


Evaluate the integral by using substitution.

`int_0^(pi/2) sqrt(sin phi) cos^5 phidphi`


Evaluate the integral by using substitution.

`int_0^1 sin^(-1) ((2x)/(1+ x^2)) dx`


Evaluate the integral by using substitution.

`int_0^2 xsqrt(x+2)`  (Put x + 2 = `t^2`)


The value of the integral `int_(1/3)^4 ((x- x^3)^(1/3))/x^4` dx is ______.


If `f(x) = int_0^pi t sin  t  dt`, then f' (x) is ______.


`int 1/(1 + cos x)` dx = _____

A) `tan(x/2) + c`

B) `2 tan (x/2) + c`

C) -`cot (x/2) + c`

D) -2 `cot (x/2)` + c


Evaluate of the following integral: 

\[\int x^\frac{5}{4} dx\]

Evaluate of the following integral: 

\[\int 3^x dx\]

Evaluate of the following integral:

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

Evaluate of the following integral:

\[\int 3^{2 \log_3} {}^x dx\]

Evaluate: 

\[\int\frac{1}{a^x b^x}dx\]

Evaluate:

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

Evaluate: 

\[\int\frac{2 \cos^2 x - \cos 2x}{\cos^2 x}dx\]

Evaluate:

\[\int\frac{e\log \sqrt{x}}{x}dx\]

Evaluate the following integral:

\[\int\limits_{- 4}^4 \left| x + 2 \right| dx\]

Evaluate the following integral:

\[\int\limits_{- 2}^2 \left| x + 1 \right| dx\]

 


Evaluate the following integral:

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

 


Evaluate the following integral:

\[\int\limits_0^4 \left( \left| x \right| + \left| x - 2 \right| + \left| x - 4 \right| \right) dx\]

Evaluate each of the following integral:

\[\int_\frac{\pi}{6}^\frac{\pi}{3} \frac{\sqrt{\sin x}}{\sqrt{\sin x} + \sqrt{\cos x}}dx\]

 


Evaluate each of the following integral:

\[\int_{- \frac{\pi}{3}}^\frac{\pi}{3} \frac{1}{1 + e^\ tan\ x}dx\]

 


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

Evaluate the following integral:

\[\int_0^\frac{\pi}{2} \frac{\tan^7 x}{\tan^7 x + \cot^7 x}dx\]

Evaluate the following integral:

\[\int_{- 2}^2 \frac{3 x^3 + 2\left| x \right| + 1}{x^2 + \left| x \right| + 1}dx\]

Evaluate the following integral:

\[\int_0^\pi \left( \frac{x}{1 + \sin^2 x} + \cos^7 x \right)dx\]

Evaluate the following integral:

\[\int_0^\frac{\pi}{2} \frac{a\sin x + b\sin x}{\sin x + \cos x}dx\]

 


Evaluate : \[\int\limits_{- 2}^1 \left| x^3 - x \right|dx\] .


Evaluate: `int_  e^x ((2+sin2x))/cos^2 x dx`


Find: `int_  (3"x"+ 5)sqrt(5 + 4"x"-2"x"^2)d"x"`.


Find: `int (dx)/sqrt(3 - 2x - x^2)`


The value of `int_0^1 (x^4(1 - x)^4)/(1 + x^2) dx` is


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


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×