A Note In order to quicken this method, we can proceed as follows: After performing steps , and , there is no need of step . Here, the integral will be kept in the new variable itself, and the limits of the integral will accordingly be changed, so that we can perform the last step. Let us illustrate this by examples. Example Evaluate .
Solution Put t = x + , then dt = x dx . t dt t = ( x + Hence, – ( ( ) – (– ) ( Alternatively , first we transform the integral and then evaluate the transformed integral with new limits. Let t = x + . Then dt = x dx .
Note that, when x = – , t = and when x = , t = Thus, as x varies from – to , t varies from to Therefore t dt – = ( Example Evaluate – INTEGRALS Solution Let t = tan – x , then . The new limits are, when x = , t = and when x = , . Thus, as x varies from to , t varies from to π . Therefore – t dt – EXERCISE .
Evaluate the integrals in Exercises to using substitution. . x + . φ φ φ .
. e dx Choose the correct answer in Exercises and . . The value of the integral is (A) (B) (C) (D) .
If f ( x ) = sin x t t dt , then f ′ ( x ) is (A) cos x + x sin x (B) x sin x (C) x cos x (D) sin x +