मराठी

Evaluate the following: d∫(cos5x+cos4x)1-2cos3xdx

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प्रश्न

Evaluate the following:

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

बेरीज
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उत्तर

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

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

= `int (2cos  (9x)/2 * cos  x/2)/(1 - 4 cos^2  (3x)/2 + 2) "d"x`

= `int (2cos  (9x)/2 * cos  x/2)/(3 - 4 cos^2  (3x)/2)  "d"x`

= `- int (2 cos  (9x)/2 * cos  x/2)/(4 cos^2  (3x)/2 - 3)  "d"x`

= `- int (2cos  (9x)/2 * cos  x/2 * cos  (3x)/2)/(4 cos^2  (3x)/2 - 3 cos  (3x)/2) "d"x`  ....`["Multiplying and dividing by" cos  (3x)/2]`

= `int (2  cos  (9x)/2 * cos  x/2 * cos  (3x)/2)/(cos 3 * (3x)/2)  "dx"`  ......[∵ cos 3x = 4 cos3x – 3 cos x]

= `- int (2cos  (9x)/2 * cos  x/2 * cos  (3x)/2)/(cos  (9x)/2)  "d"x`

= `- int 2 cos  (3x)/2 * cos  x/2  "d"x`

= `- int [cos((3x)/2 + x/2) + cos((3x)/2 - x/2)] "d"x`

= `- int (cos 2x + cos x) "d"x`  ....[∵ 2 cos A cos B = cos (A + B) + cos (A – B)]

= `- int cos 2x  "d"x - int cos x "d"x`

= `- 1/2 sin 2x - sin x + "C"`

Hence, I = `- [1/2 sin 2x + sin x] + "C"`.

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पाठ 7: Integrals - Exercise [पृष्ठ १६४]

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