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Find d∫01x(tan-1x) dx

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प्रश्न

Find `int_0^1 x(tan^-1x)  "d"x`

योग
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उत्तर

I = `int_0^1x(tan^-1x)^2  "d"x`

Integrating by parts, we have

I = `x^2/2[(tan^-1x)^2]_0^1 - 1/2 int_0^1 x^2 * 2 (tan^-1x)/(1 + x^2)  "d"x`

= `pi^2/32 - int_0^1  x^2/(1 + x) * tan^-1  x"d"x`

= `pi^2/32 - 1_1`, where I1 = `int_0^1 x^2/(1 + x^2) tan^-1 x"d"x`

Now I1 = `int_0^1 (x^2 + 1 - 1)/(1 + x^2) tan^-1x "d"x`

= `int_0^1 tan^-1 x"d"x - int_0^1 1/(1 + x^2) tan^-1 x"d"x`

= `"I"_2 - 1/2 ((tan^-1x)^2)_0^1`

= `"I"_2 - pi^2/32`

Here I2 = `int_0^1 tan^-1 x"d"x = (x tan^-1x)_0^1 - int_0^1 x/(1 + x^2)  "d"x`

= `pi/4 - 1/2(log|1 + x^2|)_0^1`

= `pi/4 - 1/2 log2`

Thus I2 = `pi/4 - 1/2 log 2 - pi^2/32`

Therefore, I = `pi^2/32 - pi/4 + 1/2 log2 + pi^2/32`

= `pi^2/16 - pi/4 + 1/2 log2`

= `(pi^2 - 4pi)/16 + log sqrt(2)`.

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अध्याय 7: Integrals - Solved Examples [पृष्ठ १५६]

APPEARS IN

एनसीईआरटी एक्झांप्लर Mathematics Exemplar [English] Class 12
अध्याय 7 Integrals
Solved Examples | Q 18 | पृष्ठ १५६

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