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Prove that: int sqrt(a^2 – x^2) * dx = x/2 * sqrt(a^2 – x^2) + a^2/2 * sin^–1(x/a) + c

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प्रश्न

Prove that: `int sqrt(a^2 - x^2) * dx = x/2 * sqrt(a^2 - x^2) + a^2/2 * sin^-1(x/a) + c`

प्रमेय
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उत्तर

Let I = `int sqrt(a^2 - x^2) dx`

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

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

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

= `sqrt(a^2 - x^2)*x - int 1/(2sqrt(a^2 - x^2))(0 - 2x)*x  dx`

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

= `xsqrt(a^2 - x^2) - int (a^2 - x^2 - a^2)/sqrt(a^2 - x^2)dx`

= `xsqrt(a^2 - x^2) - int sqrt(a^2 - x^2)dx + a^2 int dx/sqrt(a^2 - x^2)`

= `xsqrt(a^2 - x^2) - I + a^2sin^-1(x/a) + c_1`

∴ 2I = `xsqrt(a^2 - x^2) + a^2sin^-1(x/a) + c_1`

∴ I = `x/2 sqrt(a^2 - x^2) + a^2/2 sin^-1(x/a) + c_1/2`

∴ `int sqrt(a^2 - x^2)dx = x/2 sqrt(a^2 - x^2) + a^2/2sin^-1(x/a) + c`, where `c = c_1/2`.

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2015-2016 (March)

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