Definite Integration by Reduction Methods Consider

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Definite Integration by Reduction Methods Definite Integration by Reduction Methods

Consider  2 0 cos xdx. I n n 21 )1( cossin 1, nn n IConsider 2 0 cos xdx. I n n 21 )1( cossin 1, nn n I nn xx n. I Now 22 012 0 1 cossin 1 cos, nnn n I nn xx nxdx. ISo

Consider  2 0 cos xdx. I n n 2 21 211 1 , 1 0Consider 2 0 cos xdx. I n n 2 21 211 1 , 1 0 cos 01 0 2 sin 1 1 0 cos 0 sin 1 2 cos 2 sin 1 nn nnn n I nn IHence I nn nn. I

 • Using exactly the same method obtains exactly the same result.  • Hence, Now • Using exactly the same method obtains exactly the same result. • Hence, Now consider 2 0 sin xdx. I n n 2 2 0 1 sincos nn nn I n n Ixdxxdx

Using this reduction formula successively • Repeatedly applying the formula, we find that: . 4 5Using this reduction formula successively • Repeatedly applying the formula, we find that: . 4 5 2 31 1 642 etc I n n n I I n n I nn nn nn

Using this reduction formula successively • The final term will depend on if n is evenUsing this reduction formula successively • The final term will depend on if n is even or odd. • If n is even: • Also: • Hence in both cases, when n is even: 02 1 4 3. . . 2 31 I n n In 2 cos 2 0 0 0 xdx. I 2 sin 2 0 0 0 xdx. I 22 1 4 3. . . 2 31 n n I n

Using this reduction formula successively • If n is odd:  • Also:  • HenceUsing this reduction formula successively • If n is odd: • Also: • Hence in both cases, when n is odd: 13 2 5 4. . . 2 31 I n n In 1 cos 2 0 1 1 xdx. I 1 sin 2 0 1 1 xdx. I 3 2 5 4. . . 2 31 n n I n

Example • Evaluate • n is even so,  • Hence, 22 1 4 3 6Example • Evaluate • n is even so, • Hence, 22 1 4 3 6 5 8 7 268 78 48 58 28 38 8 18 88 I I 2 0 8 cos xdx 256 35 8 I

Example • Evaluate • n is odd so,  • Hence,     Example • Evaluate • n is odd so, • Hence, 32 5 4 7 6 47 57 27 37 7 17 77 II 2 0 7 sin xdx 35 16 7 I