हिंदी

Evaluate the following integrals : ∫9-xx.dx

Advertisements
Advertisements

प्रश्न

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

योग
Advertisements

उत्तर

 Let I = `int sqrt((9 - x)/x).dx`

= `int sqrt((9 - x)/x.(9 - x)/(9 - x)).dx`

= `int (9 - x)/sqrt(9x - x^2).dx`

Let 9 – x = `"A"[d/dx (9x - x^2)] + "B"`

= A(9 – 2x) + B
∴ 9 – x = (9A + B) – 2Ax
Comparing the coefficient of x and constant on both the sides, we get
– 2A = – 1 and 9A + B = 9

∴ `"A" = (1)/(2) and 9(1/2) + "B"` = 9

∴ B = `(9)/(2)`

∴ 9 – x = `(1)/(2)(9 - 2x) + (9)/(2)`

∴ I = `int (1/2(9 - 2x) + 9/2)/sqrt(9x - x^2).dx`

= `(1)/(2) int (9 - 2x)/sqrt(9x - x^2).dx + (9)/(2) int (1)/sqrt(9x - x^2).dx`

= `(1)/(2)"I"_1 + (9)/(2)"I"_2`

In I1, put 9x – x2 = t
∴ (9 – 2x)dx = dt

∴ I1 = `int (1)/sqrt(t)dt`

= `intt^(-1/2)dt`

= `t^(1/2)/(1/2) + c_1`

= `2sqrt(9x - x^2) + c_1`

I2 = `int(1)/sqrt(81/4 - (x^2 - 9x + 81/4)).dx`

= `int (1)/sqrt((9/2)^2 - (x - 9/2)^2).dx`

= `sin^-1((x - 9/2)/(9/2)) + c_2`

== `sin^-1((2x - 9)/9) + c_2`

∴ I = `sqrt(9x - x^2) + (9)/(2) sin^-1((2x - 9)/9) + c`, where c = c1 + c2.

shaalaa.com
  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 3: Indefinite Integration - Exercise 3.2 (C) [पृष्ठ १२८]

APPEARS IN

बालभारती Mathematics and Statistics 2 (Arts and Science) [English] Standard 12 Maharashtra State Board
अध्याय 3 Indefinite Integration
Exercise 3.2 (C) | Q 1.7 | पृष्ठ १२८

वीडियो ट्यूटोरियलVIEW ALL [2]

संबंधित प्रश्न

Prove that `int_a^bf(x)dx=f(a+b-x)dx.` Hence evaluate : `int_a^bf(x)/(f(x)+f(a-b-x))dx`


Find the particular solution of the differential equation x2dy = (2xy + y2) dx, given that y = 1 when x = 1.


Evaluate : `∫1/(3+2sinx+cosx)dx`


\[\int e^x \sqrt{e^{2x} + 1} \text{ dx}\]

\[\int\sqrt{16 x^2 + 25} \text{ dx}\]

\[\int\sqrt{2 x^2 + 3x + 4} \text{ dx}\]

Write a value of

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

 Write a valoue of \[\int \sin^3 x \cos x\ dx\]

 


Write a value of\[\int\frac{\sin x + \cos x}{\sqrt{1 + \sin 2x}} dx\]


Write a value of\[\int e^{ax} \sin\ bx\ dx\]


Write a value of\[\int e^{ax} \cos\ bx\ dx\].

 


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

 

 


\[\int\frac{\cos^5 x}{\sin x} \text{ dx }\]

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

Evaluate the following integrals : `int (sin2x)/(cosx)dx`


Integrate the following functions w.r.t. x : `(e^(2x) + 1)/(e^(2x) - 1)`


Integrate the following functions w.r.t.x:

`(2sinx cosx)/(3cos^2x + 4sin^2 x)`


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


Integrate the following functions w.r.t. x : `(sinx + 2cosx)/(3sinx + 4cosx)`


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


Integrate the following functions w.r.t. x : `int (1)/(3 + 2sinx).dx`


Integrate the following functions w.r.t. x : `int (1)/(3 + 2sin x - cosx)dx`


Integrate the following functions w.r.t. x : `int (1)/(cosx - sinx).dx`


Evaluate the following integrals : `int sqrt((x - 7)/(x - 9)).dx`


Integrate the following with respect to the respective variable : `(x - 2)^2sqrt(x)`


Integrate the following w.r.t.x: `(3x + 1)/sqrt(-2x^2 + x + 3)`


Evaluate `int 1/(x (x - 1))` dx


Evaluate the following.

`int "x" sqrt(1 + "x"^2)` dx


Evaluate the following.

`int "x"^5/("x"^2 + 1)`dx


Evaluate:

`int (5x^2 - 6x + 3)/(2x − 3)` dx


Evaluate `int 1/((2"x" + 3))` dx


Evaluate: If f '(x) = `sqrt"x"` and f(1) = 2, then find the value of f(x).


Evaluate: `int 1/(sqrt("x") + "x")` dx


Evaluate: `int (2"e"^"x" - 3)/(4"e"^"x" + 1)` dx


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


State whether the following statement is True or False:

`int sqrt(1 + x^2) *x  "d"x = 1/3(1 + x^2)^(3/2) + "c"`


`int (1 + x)/(x + "e"^(-x))  "d"x`


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


Evaluate `int1/(x(x - 1))dx`


Evaluate.

`int (5x^2-6x+3)/(2x-3)dx`


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


The value of `int ("d"x)/(sqrt(1 - x))` is ______.


Evaluate the following.

`int (x^3)/(sqrt(1 + x^4)) dx`


Evaluate `int1/(x(x-1))dx` 


Evaluate the following.

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


Evaluate `int (5x^2 - 6x + 3)/(2x - 3) dx`


Evaluate:

`intsqrt(sec  x/2 - 1)dx`


What is \[\int\cot t\,dt\] in the evaluation of \[\int\frac{\sin x}{\sin(x+a)}\,dx\]?


Integration by substitution is the reverse process of which rule?


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×