{"id":3096,"date":"2021-07-18T12:52:40","date_gmt":"2021-07-18T12:52:40","guid":{"rendered":"https:\/\/mathemerize.com\/?p=3096"},"modified":"2021-11-26T01:35:59","modified_gmt":"2021-11-25T20:05:59","slug":"formula-for-integration-by-parts","status":"publish","type":"post","link":"https:\/\/mathemerize.com\/formula-for-integration-by-parts\/","title":{"rendered":"What is the Formula for Integration by Parts ?"},"content":{"rendered":"
If u and v are two functions of x, then the formula for integration by parts is –<\/p>\n
\n\\(\\int\\) u.v dx = u \\(\\int\\) v dx – \\(\\int\\)[\\(du\\over dx\\).\\(\\int\\)v dx]dx<\/p>\n<\/blockquote>\n
i.e The integral of the product of two functions = (first function) \\(\\times\\) (Integral of Second function) – Integral of { (Diff. of first function) \\(\\times\\) (Integral of Second function)}<\/p>\n
Note<\/strong> –\u00a0 We can choose the first function as the function which comes first in the word ILATE,\u00a0<\/strong>where\u00a0<\/p>\n
\nI – Stands for the Inverse Trigonometric Function<\/p>\n
L – Stands for the Logarithmic Function<\/p>\n
A – Stands for the Algebraic Function<\/p>\n
T – Stands for the Trigonometric Function<\/p>\n
E – Stands for the Exponential Function<\/p>\n<\/blockquote>\n\n\n
Example : <\/span> Solve the integral \\(\\int\\) \\(x sin3x\\) dx using the formula for integration by parts.<\/p>\n
Solution : <\/span>Here both the functions viz. x and sin3x are easily integrable and the derivative of x is one, a less complicated function. Therefore, we take x as the first function and sin3x as the second function.
\n\\(\\therefore\\) I = \\(\\int\\) x cos3x dx
\n= x [\\(\\int\\) sin3x dx] – \\(\\int\\)[\\({d\\over dx}(x)\\) \\(\\times\\) \\(\\int\\)sin3x dx] dx
\n= x (\\(-1\\over 3\\) cos3x) – \\(\\int\\) [\\(-1\\over 3\\) cos3x] dx
\n\\(\\implies\\) I = \\(-1\\over 3\\) xcos3x + \\(1\\over 3\\) \\(\\int\\) cos3x dx
\n\\(\\implies\\) I = -\\(1\\over 3\\) xcos3x + \\(1\\over 9\\) sin3x + C
\n<\/p>\n\n\n\nExample : <\/span> Solve the integral \\(\\int\\) \\(x sin^{-1}x\\) dx using the formula for integration by parts.<\/p>\n
Solution : <\/span>Taking \\(sin^{-1}x\\) as the first function and x as the second function by using the ILATE rule.
\nI = \\(\\int\\) \\(x sin^{-1}x\\)
\n= (\\(sin^{-1}x\\))\\(x^2\\over 2\\) – \\(\\int\\)[\\(1\\over \\sqrt{1-x^2}\\) \\(\\times\\) \\(x^2\\over 2\\)] dx
\n= \\(x^2\\over 2\\)(\\(sin^{-1}x\\)) + \\(1\\over 2\\) \\(\\int\\)[\\((-x)^2\\over \\sqrt{1-x^2}\\) dx = \\(x^2\\over 2\\)(\\(sin^{-1}x\\)) + \\(1\\over 2\\) \\(\\int\\)[\\({1-x^2-1}\\over \\sqrt{1-x^2}\\) dx
\n\\(\\implies\\) I = \\(x^2\\over 2\\)(\\(sin^{-1}x\\)) + \\(1\\over 2\\) {\\(\\int\\)[\\({1-x^2}\\over \\sqrt{1-x^2}\\) dx – \\(1\\over \\sqrt{1-x^2}\\) dx}
\n\\(\\implies\\) I = \\(x^2\\over 2\\)(\\(sin^{-1}x\\)) + \\(1\\over 2\\) {\\(\\int\\)[\\(\\sqrt{1-x^2}\\) dx – \\(1\\over \\sqrt{1-x^2}\\) dx}
\nI = \\(x^2\\over 2\\)(\\(sin^{-1}x\\)) + \\(1\\over 2\\) {\\({1\\over 2}x\\)[\\(\\sqrt{1-x^2}\\) dx + \\(1\\over 2\\) \\(\\sin^{-1}x\\) – \\(sin^{-1}x\\)] + C
\n\\(\\implies\\) I = \\(x^2\\over 2\\)(\\(sin^{-1}x\\)) + \\(1\\over 4\\)\\(x \\sqrt{1-x^2}\\) dx – \\(1\\over 4\\) \\(\\sin^{-1}x\\) + C
\n<\/p>\n\n\nHope you learnt what is the formula for integration by parts, learn more concepts of Indefinite Integration and practice more questions to get ahead in the competition. Good luck!<\/p>\n\n\n