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Integrate the function in x cos-1 x.

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प्रश्न

Integrate the function in x cos-1 x.

योग
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उत्तर

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

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

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

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

∴ `I = x^2/2 cos^-1 x+ 1/2 I_1`             ....(i)

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

Put x = cos θ 

⇒ dx = -sinθ dθ 

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

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

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

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

`= - 1/2 (theta + cos theta sqrt (1 - cos^2 theta)) + C`

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

From (i) and (ii), we get

`I = (2x^2 - 1) (cos^-1 x)/4 - x/4 sqrt (1 - x^2) + C`

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अध्याय 7: Integrals - Exercise 7.6 [पृष्ठ ३२७]

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एनसीईआरटी Mathematics Part 1 and 2 [English] Class 12
अध्याय 7 Integrals
Exercise 7.6 | Q 9 | पृष्ठ ३२७

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