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Find the integrals of the following: 19+8x-x2 - Mathematics

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प्रश्न

Find the integrals of the following:

`1/sqrt(9 + 8x - x^2)`

योग
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उत्तर

`int 1/sqrt(9 + 8x - x^2)  "d"x = int ("d"x)/sqrt(9 - (x^2 - 8x))`

= `int ("d"x)/sqrt(9 - [(x - 4)^2 - 4^2]`

= `int ("d"x)/sqrt(9 - [(x - 4)^2 - 16]`

= `int ("d"x)/sqrt(9 - (x - 4)^2 + 16)`

= `int ("d"x)/sqrt(25 - (x - 4)^2`

= `int ("d"x)/sqrt(5^2 - (x - 4)^2`

Put x – 4 = t

dx = dt

`int ("d")/sqrt(9 + 8x - x^2) = int ("d"x)/sqrt(5^2 - "t"^2)`

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

`int ("d"x)/sqrt(9 + 8x - x^2) = sin^-1 ("t"/5) + "c"`

= `sin^-1 ((x - 4)/5) + "c"`

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  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 11: Integral Calculus - Exercise 11.10 [पृष्ठ २१९]

APPEARS IN

सामाचीर कलवी Mathematics - Volume 1 and 2 [English] Class 11 TN Board
अध्याय 11 Integral Calculus
Exercise 11.10 | Q 3. (iii) | पृष्ठ २१९

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