{"id":5837,"date":"2021-10-03T19:53:50","date_gmt":"2021-10-03T14:23:50","guid":{"rendered":"https:\/\/mathemerize.com\/?p=5837"},"modified":"2021-11-25T23:12:38","modified_gmt":"2021-11-25T17:42:38","slug":"integration-of-sin-inverse-x","status":"publish","type":"post","link":"https:\/\/mathemerize.com\/integration-of-sin-inverse-x\/","title":{"rendered":"Integration of Sin Inverse x"},"content":{"rendered":"
Here you will learn proof of integration of sin inverse x or arcsin x and examples based on it.<\/p>\n
Let’s begin –<\/p>\n
\nThe integration of sin inverse x or arcsin x is \\(xsin^{-1}x\\) + \\(\\sqrt{1 – x^2}\\) + C<\/strong><\/p>\n
Where C is the integration constant.<\/p>\n
i.e. \\(\\int\\) \\(sin^{-1}x\\) = \\(xsin^{-1}x\\) + \\(\\sqrt{1 – x^2}\\) + C<\/p>\n<\/blockquote>\n
Proof :\u00a0<\/strong><\/p>\n
\nWe have, I = \\(\\int\\) \\(sin^{-1}x\\) dx<\/p>\n
Let \\(sin^{-1}x\\) = t,<\/p>\n
Then, x = sint\u00a0<\/p>\n
\\(\\implies\\) dx = d(sin t) = cos t dt<\/p>\n
\\(\\therefore\\) I =\u00a0\\(\\int\\) \\(sin^{-1}x\\) dx<\/p>\n
\\(\\implies\\) I = \\(\\int\\) t cost dt<\/p>\n
By using integration by parts formula,<\/p>\n
I = t sin t – \\(\\int\\) 1. (sin t) dt<\/p>\n
I = t sin t – \\(\\int\\) sint dt<\/p>\n
= t sin t + cos t + C<\/p>\n
= t sin t + \\(\\sqrt{1 – sin^2 t}\\) + C<\/p>\n
Now, Put t = \\(sin^{-1}x\\) here<\/p>\n
\\(\\implies\\) I = x \\(sin^{-1}x\\) + \\(\\sqrt{1 – x^2}\\) + C<\/p>\n
Hence, \\(\\int\\) \\(sin^{-1}x\\) dx = x \\(sin^{-1}x\\) + \\(\\sqrt{1 – x^2}\\) + C<\/p>\n<\/blockquote>\n
Example<\/span><\/strong> : Evaluate \\(\\int\\) \\(x sin^{-1} x\\) dx<\/p>\n
Solution<\/strong><\/span> : We have,<\/p>\n
I = \\(\\int\\)\u00a0 \\(x sin^{-1} x\\) dx<\/p>\n
By using integration by parts formula<\/a>,<\/p>\n
I = \\(sin^{-1} x\\) \\(x^2\\over 2\\) – \\(\\int\\) \\(1\\over \\sqrt{1 – x^2}\\) \\(\\times\\) \\(x^2\\over 2\\) dx<\/p>\n
I =\u00a0 \\(x^2\\over 2\\) \\(sin^{-1} x\\) + \\(1\\over 2\\) \\(\\int\\) \\(-x^2\\over \\sqrt{1 – x^2}\\) dx<\/p>\n
= \\(x^2\\over 2\\) \\(sin^{-1} x\\) + \\(1\\over 2\\) \\(\\int\\) \\(1 – x^2 – 1\\over \\sqrt{1 – x^2}\\)\u00a0 dx<\/p>\n
= \\(x^2\\over 2\\) \\(sin^{-1} x\\) + \\(1\\over 2\\) { \\(\\int\\) \\(1 – x^2\\over \\sqrt{1 – x^2}\\) – \\(\\int\\) \\(1\\over \\sqrt{1 -x^2}\\) } dx<\/p>\n
\\(\\implies\\) I = \\(x^2\\over 2\\) \\(sin^{-1} x\\) + \\(1\\over 2\\) { \\(\\int\\) \\(\\sqrt{1 – x^2}\\) – \\(\\int\\) \\(1\\over \\sqrt{1 -x^2}\\) } dx<\/p>\n
By using integration formula<\/a> of \\(\\sqrt{a^2 – x^2}\\),<\/p>\n
\\(\\implies\\) I = \\(x^2\\over 2\\) \\(sin^{-1} x\\) + \\(1\\over 2\\) [{ \\(1\\over 2\\) \\(x\\sqrt{1 – x^2}\\) – \\(1\\over 2\\) \\(sin^{-1} x\\) } – \\(sin^{-1} x\\) ] + C<\/p>\n
\\(\\implies\\) I = \\(x^2\\over 2\\) \\(sin^{-1} x\\) +\u00a0 \\(1\\over 4\\) \\(x\\sqrt{1 – x^2}\\) – \\(1\\over 4\\) \\(sin^{-1} x\\) + C<\/p>\n
\nRelated Questions<\/h3>\n
What is the Differentiation of sin inverse x ?<\/a><\/p>\n
What is the Integration of sin x ?<\/a><\/p>\n
What is the integration of sin inverse root x ?<\/a><\/p>\n\n\n