Question

Find the\(\rm \smallint si{n^{ - 1}}x\;dx\)
1. \(x\;co{s^{ - 1}}x\; + \;\sqrt {1 - {x^2}} + c\)
2. \(x\;si{n^{ - 1}}x\; - \;\sqrt {1 - {x^2}} + c\)
3. \(x\;co{s^{ - 1}}x\; - \;\sqrt {1 - {x^2}} + c\)
4. \(x\;si{n^{ - 1}}x\; + \;\sqrt {1 - {x^2}} + c\)

Answer 1

Correct Answer - Option 4 : \(x\;si{n^{ - 1}}x\; + \;\sqrt {1 - {x^2}} + c\)

Concept:

Integration by parts

Using ILATE rules

I → Inverse function

L → Log function

A → Algebraic function

T → Trigonometry function

E → Exponential function

\(\rm \smallint u.v\;dx = u\;\smallint v\;dx - [\;\smallint (\frac{{du}}{{dx}})\;\smallint v\;dx\;]\;dx\)

Calculation:

\(\rm I = \smallint si{n^{ - 1}}x\;dx\)

= \( si{n^{ - 1}}x.\;\smallint 1\;dx - [\;\smallint \frac{{d\left( {si{n^{ - 1}}x} \right)}}{{dx}}\;\;\smallint 1\;dx\;]\;dx\)

= \( si{n^{ - 1}}x.\;\smallint 1\;dx - [\;\smallint \frac{{d\left( {si{n^{ - 1}}x} \right)}}{{dx}}\;\;(\smallint 1\;dx)]\;dx\)

= \( x.\;si{n^{ - 1}}x - [\smallint \frac{1}{{\sqrt {\left( {1 - {x^2}} \right)} }}\;.\;x\;]\;dx\)

Put 1 – x² = t²

⇒ -2x dx = 2t dt

⇒ x dx = -t dt

Now, I =  \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaaeaaaaaaaaa8 % qacaWG4bGaaiOlaiaacckacaWGZbGaamyAaiaad6gapaWaaWbaaSqa % beaapeGaeyOeI0IaaGymaaaakiaadIhacqGHsislcqGHRiI8daWcaa % WdaeaapeGaaGymaaWdaeaapeWaaOaaa8aabaWdbiaadshapaWaaWba % aSqabeaapeGaaGOmaaaaaeqaaaaakmaabmaapaqaa8qacqGHsislca % WG0baacaGLOaGaayzkaaGaaiiOaiaadsgacaWG0baaaa!4B4A! x.\;si{n^{ - 1}}x - \smallint \frac{1}{{\sqrt {{t^2}} }}\left( { - t} \right)\;dt\)\(x.\;si{n^{ - 1}}x - \smallint \frac{1}{{\sqrt {{t^2}} }}\left( { - t} \right)\;dt\)

= \(x.si{n^{ - 1}}x + \;\smallint 1\;dt\)

= \(\rm x.si{n^{ - 1}}x + \;t\) + c

Putting the value of t ⇒ \(t = \;\sqrt {1 - {x^2}} \)

I = \(x\;si{n^{ - 1}}x\; + \;\sqrt {1 - {x^2}} + c\)