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Moreover, it is not possible to convert it to a standard form by using algebraic operations and/or trigonometric identities. However, it is possible to convert it into a standard form as follows: We put sin x ¼ t and differentiate both sides of this equation with respect to t to obtain cos x dx Ð ¼ dt. Now, by using Ð these relations in the expression for the element of integration, we get sin3 x cos x dx ¼ t3 dt, which can be easily evaluated. We have ð t3 dt ¼ t4 sin4 x þc þc¼ 4 4 Note that, in the process of converting the above integrand into a standard form, we had to change the variable of integration from x to t.

Rohde, G. C. Jain, Ajay K. Poddar, and A. K. Ghosh. Ó 2012 John Wiley & Sons, Inc. Published 2012 by John Wiley & Sons, Inc. 43 44 INTEGRATION BY SUBSTITUTION: CHANGE OF VARIABLE OF INTEGRATION Example (1): Find Ð 25 (a) ðx3 þ 2xÞ ð3x2 þ 2Þdx Ð 12 (b) sin x cos x dx Solution: Ð 25 (a) To evaluate ðx3 þ 2xÞ ð3x2 þ 2Þdx, we observe that d 3 (x þ 2x) ¼ 3x2 þ 2. dx Let f(x) ¼ x3 þ 2x, ) f 0 (x) ¼ 3x2 þ 2. Thus, by the above theorem, ð ð 25 ðx3 þ 2xÞ ð3x2 þ 2Þdx ¼ ½ f ðxÞ25 Á f 0 ðxÞdx ½ f ðxÞ26 : þc 26 26 3 ðx þ 2xÞ þc ¼ 26 ¼ Ð (b) To evaluate sin12 x cos x dx, we observe that (d/dx)(sin x) ¼ cos x.

12. 13. 14. Differentiation Formulas d Already Known to us ½ f ðxÞ ¼ f 0 ðxÞ dx d À À1 Á 1 sin x ¼ pﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ dx 1 À x2 d À À1 Á À1 cos x ¼ pﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ dx 1 À x2 d À À1 Á 1 tan x ¼ dx 1 þ x2 d À À1 Á À1 cot x ¼ dx 1 þ x2 d À À1 Á 1 sec x ¼ pﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ dx x Á x2 À 1 Á d À À1 cosecÀ1 x ¼ pﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ dx x Á x2 À 1 Corresponding Formulas for Ð Indeﬁnite Integrals f 0 ðxÞdx ¼ f ðxÞ þ c ð ð ð 8 À1 > < sin x þ c dx pﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ ¼ or 1 À x2 > : ÀcosÀ1 x þ c 8 À1 > < tan x þ c dx ¼ or 1 þ x2 > : ÀcotÀ1 x þ c 8 À1 > < sec x þ c dx pﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ ¼ or x Á x2 À 1 > : ÀcosecÀ1 x þ c 10 ANTIDERIVATIVE(S) [OR INDEFINITE INTEGRAL(S)] form, evaluating the integral depends only on recognizing the form and remembering the table of integrals.