मराठी

Integrate the function in x sin−1 x. - Mathematics

Advertisements
Advertisements

प्रश्न

Integrate the function in x sin−1 x.

बेरीज
Advertisements

उत्तर

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
  या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
पाठ 7: Integrals - Exercise 7.6 [पृष्ठ ३२७]

APPEARS IN

एनसीईआरटी Mathematics Part 1 and 2 [English] Class 12
पाठ 7 Integrals
Exercise 7.6 | Q 7 | पृष्ठ ३२७

व्हिडिओ ट्यूटोरियलVIEW ALL [2]

संबंधित प्रश्‍न

Prove that:

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


Integrate the function in `x^2e^x`.


Integrate the function in xlog x.


Integrate the function in e2x sin x.


Evaluate the following:

`int x tan^-1 x . dx`


Evaluate the following : `int x^2tan^-1x.dx`


Evaluate the following : `int(sin(logx)^2)/x.log.x.dx`


Evaluate the following:

`int x.sin 2x. cos 5x.dx`


Integrate the following functions w.r.t. x : `sqrt(5x^2 + 3)`


Integrate the following functions w.r.t. x : `sqrt((x - 3)(7 - x)`


Integrate the following functions w.r.t. x : `sec^2x.sqrt(tan^2x + tan x - 7)`


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

`e^(5x).[(5x.logx + 1)/x]`


Choose the correct options from the given alternatives :

`int (x- sinx)/(1 - cosx)*dx` =


Choose the correct options from the given alternatives :

`int sin (log x)*dx` =


Choose the correct options from the given alternatives :

`int [sin (log x) + cos (log x)]*dx` =


Integrate the following with respect to the respective variable : `(3 - 2sinx)/(cos^2x)`


Integrate the following with respect to the respective variable : cos 3x cos 2x cos x


Evaluate: `int "dx"/("x"[(log "x")^2 + 4 log "x" - 1])`


Evaluate:

∫ (log x)2 dx


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


`int sqrt(tanx) + sqrt(cotx)  "d"x`


`int 1/x  "d"x` = ______ + c


`int 1/(x^2 - "a"^2)  "d"x` = ______ + c


Evaluate `int 1/(x(x - 1))  "d"x`


Evaluate `int 1/(x log x)  "d"x`


`int logx/(1 + logx)^2  "d"x`


∫ log x · (log x + 2) dx = ?


If `int(2e^(5x) + e^(4x) - 4e^(3x) + 4e^(2x) + 2e^x)/((e^(2x) + 4)(e^(2x) - 1)^2)dx = tan^-1(e^x/a) - 1/(b(e^(2x) - 1)) + C`, where C is constant of integration, then value of a + b is equal to ______.


`int(1-x)^-2 dx` = ______


Solution of the equation `xdy/dx=y log y` is ______


`int(f'(x))/sqrt(f(x)) dx = 2sqrt(f(x))+c`


Solve the following

`int_0^1 e^(x^2) x^3 dx`


Evaluate:

`int (logx)^2 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`


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


Evaluate the following.

`intx^2e^(4x)dx`


The value of `inta^x.e^x dx` equals


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×