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प्रश्न
`int_0^{1/sqrt2} (sin^-1x)/(1 - x^2)^{3/2} dx` = ______
पर्याय
`pi/4 + 1/2log2`
`pi/4 - 1/2log2`
`pi/2 + log2`
`pi/2 - log2`
MCQ
रिकाम्या जागा भरा
उत्तर
`int_0^{1/sqrt2} (sin^-1x)/(1 - x^2)^{3/2} dx` = `underline(pi/4 - 1/2log2)`.
Explanation:
Let I = `int_0^{1/sqrt2} (sin^-1x)/(1 - x^2)^{3/2} dx`
Put `sin^-1x = t ⇒ 1/sqrt(1 - x^2)dx = dt`
∴ `I = int_0^{pi/4} t sec^2t dt`
= `[t . tant]_0^{pi/4} - int_0^{pi/4}1 . tan t dt`
= `(pi/4 . tan pi/4 - 0) - [log|sec t|]_0^{pi/4}`
= `pi/4 (1) - [log|sec pi/4| - log|sec 0|]`
= `pi/4 - (logsqrt2 - log 1)`
= `pi/4 - (log2^{1/2} - 0)`
∴ I = `pi/4 - 1/2log2`
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