English

Integrate the function in x sin−1 x.

Advertisements
Advertisements

Question

Integrate the function in x sin−1 x.

Sum
Advertisements

Solution

Let `I = int x sin^-1 x dx = int sin^-1 x* x dx`

`= sin^-1 x* (x^2/2) - int [d/dx (sin^-1 x) * x^2/2]  dx`

`= sin^-1 x (x^2/2) - int 1/sqrt (1 - x^2)* x^2/2  dx`

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

`= x^2/2 sin^-1 x - 1/2 I_1`

`I = x^2/2 sin^-1 x - 1/2 I_1`              ....(i)

Where `I_1 = int x^2/sqrt (1 - x^2)  dx`

Put x = sin θ 

⇒ dx = cosθ dθ

∴ `I_1 = int (sin^2 theta)/sqrt (1- sin^2 theta) cos d theta`

`= int (sin^2 theta)/(cos theta) * cos theta d theta`

`= int sin^2 theta d theta  = 1/2 int (1 - cos 2 theta) d theta`

`= 1/2int d theta - 1/2 int cos 2 theta d theta 1/2 theta - 1/2 (sin 2 theta)/2 + C`

  `1/2 theta - 1/2 sin theta cos theta + C`

`1/2 sin^-1x - 1/2x sqrt(1 - x^2) + C`                 ....(ii)

`[∵ sin theta = x ⇒ cos theta = sqrt (1 - sin^2 theta) = sqrt (1 - x^2)]`

From (i) and (ii), we get

∴ `I = x^2/2 sin^-1 x - 1/2 [1/2 sin^-1 x - 1/2 x sqrt(1 - x^2)] + C`

`= 1/4 sin^-1 x* (2x^2 - 1) + (x sqrt (1 - x^2))/4 + C`

shaalaa.com
  Is there an error in this question or solution?
Chapter 7: Integrals - Exercise 7.6 [Page 327]

APPEARS IN

NCERT Mathematics Part 1 and 2 [English] Class 12
Chapter 7 Integrals
Exercise 7.6 | Q 7 | Page 327

RELATED QUESTIONS

Evaluate `int_0^(pi)e^2x.sin(pi/4+x)dx`


Integrate the function in x tan-1 x.


Integrate the function in `e^x (1 + sin x)/(1+cos x)`.


Find : 

`∫(log x)^2 dx`


Evaluate the following:

`int sec^3x.dx`


Evaluate the following: `int x.sin^-1 x.dx`


Evaluate the following : `int x.cos^3x.dx`


Integrate the following functions w.r.t. x : `sqrt(4^x(4^x + 4))`


Integrate the following functions w.r.t. x: `sqrt(x^2 + 2x + 5)`.


Integrate the following functions w.r.t. x : [2 + cot x – cosec2x]e 


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


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


Choose the correct options from the given alternatives :

`int (1)/(cosx - cos^2x)*dx` =


Choose the correct options from the given alternatives :

`int cos -(3)/(7)x*sin -(11)/(7)x*dx` =


Evaluate the following.

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


Evaluate the following.

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


`int ("x" + 1/"x")^3 "dx"` = ______


Choose the correct alternative from the following.

`int (1 - "x")^(-2) "dx"` = 


Evaluate: Find the primitive of `1/(1 + "e"^"x")`


Evaluate: `int "dx"/sqrt(4"x"^2 - 5)`


`int ["cosec"(logx)][1 - cot(logx)]  "d"x`


Evaluate the following:

`int (sin^-1 x)/((1 - x)^(3/2)) "d"x`


Evaluate the following:

`int ((cos 5x + cos 4x))/(1 - 2 cos 3x) "d"x`


The value of `int_(- pi/2)^(pi/2) (x^3 + x cos x + tan^5x + 1)  dx` is


Find: `int e^x.sin2xdx`


If `π/2` < x < π, then `intxsqrt((1 + cos2x)/2)dx` = ______.


`int e^x [(2 + sin 2x)/(1 + cos 2x)]dx` = ______.


`int_0^1 x tan^-1 x  dx` = ______.


Find `int e^(cot^-1x) ((1 - x + x^2)/(1 + x^2))dx`.


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


`int (sin^-1 sqrt(x) + cos^-1 sqrt(x))dx` = ______.


Prove that `int sqrt(x^2 - a^2)dx = x/2 sqrt(x^2 - a^2) - a^2/2 log(x + sqrt(x^2 - a^2)) + c`


Complete the following activity:

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

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

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

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

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


Evaluate the following.

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


If ∫(cot x – cosec2 x)ex dx = ex f(x) + c then f(x) will be ______.


Evaluate the following.

`int x^3 e^(x^2) dx` 


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×