Solved examples integration of trigonometric functions pdf

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Solved examples integration of trigonometric functions pdf

Solved examples definite integration of trigonometric functions. Solved examples integration of trigonometric functions pdf.

3.2.1 Solve integration problems involving products and powers of sin X SIN X and COS X. COS X. 3.2.2 Solve Integration Problems involving Tan X Tan X and SEC X. s sec. 3.2.3 Use Reduction Formulas to Solve Trigonomy Integers. In this section, we analyze how to integrate a variety of trigonometic functions products. These integral are so-called

trigonometic integral. They are an important part of the technique of integration called trigonometic substitution, which is presented in trigonomic replacement. This technique allows us to convert alternative expressions that we may not be able to integrate expressions involving trigonomous functions, which we can be able to integrate the

techniques described in this section. In addition, these types of integral appear frequently when we studied polar, cylindrical and exfergious coordinate systems later. Let's start our study with Sinxsinx and Cosx.CosX products. A key view behind the strategy used to integrate combinations of products and powers from Sinxsinx and Cosxcosx involves

rewriting these expressions as sums and different integral differences ? 'sinjxcosexdx or' cosjxsinxdx. After rewriting these integral, we evaluated them using U Report. Before describing the general process in detail, let's take a look at the following examples. Evaluate ? ?Cos3xSinxdx.?? COS3XSINXDX. Use uu-substitution and leave u = cosx.u =

cosx. In this case, du = 'sinxdx.du = ?sinxdx. Thus, ?COS 3 x SIN X D x = ? ? ��? u 3 d u = 1 4 u 4 + C = ?1 4 COS 4 x + c. ? ?COS 3 x SIN X D X = ? �? U 3 D U = 1 4 U 4 + C = ?1 4 COS 4 x + C. Evaluate ? ?SIN4XCOSXDX. SIN4XCOSXDX. Evaluate ? 'cos2xsin3xdx.' cos2xsin3xdx. To convert this integral to integral form of form ? 'cosjxsinxdx cosjxsinxdx, rewrite sin3x = sin2xsinxsin3x = sin2xsinx and make substitution sin2x = 1? ?2x.sin2x = 1 products2x. Thus, ?Cos 3 x Sin 3 x D x = ?? COS 2 x (1 IC 'COS 2 x) Sin X D X Let U = COS X; Then odu = "sin xD x. =? U 2 (1? ?U 2) du = ? ? ? ? ? ? ? ?��g 2) du = 1 5 U 5?? 1 3 U 3 + C = 1 5 COS 5 x '1 3 COS 3 x + c. ? ?COS 3 x Sin 3 xDX =?

COS 2 x (1 COS 2 x) Sin XDX Let U = Cos X; then Odu = "Sin XD X. = ? ?u 2 (1?? U 2) du = ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ?��g 2) du = 1 5 u 5 ?1 3 U 3 + C = 1 5 COS 5 x '1 3 COS 3 x + C . Evaluate ? 'cos3xsin2xdx.' cos3xsin2xdx. In the next example, we see the strategy that should be applied when there are only powers of sinxsinx and

cosx.cosx. For integral of this type, identities sin2x = 12'12cos (2x) = 1? ? 'cos (2x) 2sin2x = 12x) = 1st, (2x) 2 and cos2x = 12 + 12cos (2x) = 1 + cos 2x) 2COS2X = 12 + 12COS (2x) = 1 + cos (2x) 2 are invalitive. These identities are someday known as energy reduction identities and can be derived from the angular double identity (2x) =

cos2x'sin2xcos (2x) = cos2x'sin2x and the pitagery identity COS2X + SIN2X = 1.Cas2x + SIN2X = 1. Evaluate one ? ? ? SIN2xdx. To evaluate this integral, we will use the trigonometic identity SIN2X = 12'12COS (2x) .sin2x = 12'12COS (2x). Thus, ? 'sin 2 x d x = ? ? ?? (1 2 ?��g (2 x)) d x = 1 2 x? ?1 4 Sin (2 x) + c. ? 'Sin 2 x d x = ? ?? (1 2 ? 22 (2 x))

D X = 1 2 x? �� C 1 4 Sin (2 x) + c. Evaluate ? ?Cos2xDX.?? Cos2xdx. The general process of integration of powers of Sinxsix and Cosxcosx is summarized in the following set of guidelines. To integrate ? ?COSJXSINKXDX?? COSJXSINKXDX uses the following strategies: if kk is pipe, rewrite sinkx = sink? ? '1xsinxsinkx = sink? ?' 1xsinx and use the

identity SIN2X = 1 es2xsin2x = 1?s2x to rewrite '1xsink? ? '1x in terms of cosx.cosx. Integrate using substitution u = cosx.u = cosx. This replacement becomes du = 'sinxdx.du = ICXDX. If JJ is strange, rewrite cosjx = Cosj? ? '1xcosxcosjx = Cosj? ?' 1xcosex and use the Cos2x = 1st identity, q'sin2xcos2x = 1 ?sin2x to rewrite cosmis ? '1xcosj? ?' 1x

in terms of Sinx.sinx. Integrate using the u = sinx.u = sen. This replacement becomes du = cosxdx.du = cosxdx. (Note: If both JJ and KK are strange, any of the strategies can be used a strategy or 2.) if both JJ and KK are even, use SIN2x = (1/2) A ( 1/2) COS (2x) SIN2X = (1/2) A (1/2) COS (2x) and COS2x = (1/2) + (1/2) COS (2x) .COM2X = (1/2 ) +

(1/2) COS (2x). After the applications of these formulas, simplify and strategies reapply 1 to 3, as appropriate. Evaluate a ?Cos8xsin5xdx.? ?? Cos8xSin5xdx. From SINXSINX feed, the use of use 1. Thus, one ? ?COS 8 x Sin 5 x D x = a?? COS 8 x SIN 4 x SIN X D X PARTA SIN X. = ? � ? 8 x (SIN 2 x) 2 SEN XDX sin reconfiguration 4 x = (Sin 2 x) 2. =

?, ?COS 8 x (1 to COS 2 x) 2 SEN XDX substitute SIN 2 x = 1 to COS 2 x. = ?, ? ? ? ? �� 8 (1 to U 2) 2 (a d u) is u = Cos x and d u = a sen x d x. = ?, ? ? ? ? ? ? � (u 8 + 2 u 10 a U 12) D U expand. = U1 9 9 + 2 11 U 11 to 1 13 13 U + C to assess the integral. = A 1 9 COS 9 + 2 x 11 COS 11 x to 1 13 COS 13 x + C. Replace U = COS X. ? � ? � ?COS 8

x Sin 5 x D x = A ? �? COS 8 x 4 x SIN SIN X D X PARTA SEN X. = ? � ? 8 x (SIN 2 x) 2 SEN XDX sin reconfiguration 4 x = (SIN 2 x) 2. = ?, ?COS 8 x (1 to COS 2 x) 2 SEN XDX substitute SIN 2 x = 1 to COS 2 x. = ?, ? ? ? ? �� 8 (1 to U 2) 2 (a d u) is u = Cos x and d u = a sen x d x. = ?, ? ? ? ? ? ? � (u 8 + 2 u 10 a U 12) d u expand. = U1 9 9 + 2 11

U 11 to 1 13 13 U + C to assess the integral. = A 1 9 COS 9 + 2 x 11 COS 11 x to 1 13 COS 13 x + C. Replace U = COS X. Evaluate a "sin4xdx.? �? SIN4XDX. Since SINXSINX feed is yet (k = 4) (k = 4) and Cosxcosx feed is yet (J = 0), (J = 0), we must use the strategy 3. Thus, a sin 4 xDX = ? ��, ? ?? (Sin 2 x) DX 2 Sin reconfiguration of 4 x = (Sin 2 x)

2. = A ? ? �� (1 2 to 1 2 (2x)) 2 XD substitute sin 2 x = 1 2 to 1 2 cos (2x). = ? � (1 4 1 2 Cos (2 x) + 1 4 Cos 2 (2x)) XD Expand (1 2 to 1 2 Cos (2x)) 2. = ?, ? ?(1 4 1 2 cos (2 x) + 1 4 (1 2 + 1 2 cos (4x)) dx. ?Sin 4 xDX = ? �? (SIN 2 x) 2 DX reconfiguration of sin 4 x = (SIN 2 x) 2. = A ? ? ? ? ? ? ? 1 2 COS (2 x) + 1 4 COS 2 (2x)) DX Expand (1 2 to 1 2

COS (2x)) 2. = A? ?(1 4 to 1 2 Cos (2x) + 1 4 (1 + 2 1 2 cos (4x)) DX. Since COS2 (2x) COS2 (2x) has a power still, SUBLIDATE COS2 (2x) = 12 + 12COS (4x): COS2 (2x) = 12 + 12COS ( 4x): = A ? ? � (3 8 to 1 2 cos (2 x) + 1 8 cos (4x)) DX Simplify = 3.8 XA 1 4 Sin (2x) + 1 32 Sen (x 4) + C to evaluate the integral. = A ? ? \ (3 8 to 1 2 COS (2 x) + 1 8

COS (4x)) DX Simplify. = 3 x 8 1 ? ? 4 Sin (2x) + 1 32 Sen ( x 4) + C to evaluate the integral. evaluate a ?cos3xdx.? �? COS3XDX. evaluate a ?COS2 (3x) ? �

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