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picture1_Calculus Pdf 168936 | Calc Trig Derivatives Integrals


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File: Calculus Pdf 168936 | Calc Trig Derivatives Integrals
calculus trigonometricderivativesandintegrals trigonometricderivatives d 0 d 0 d 2 0 dx sin x cos x x dx cos x sin x x dx tan x sec x x d 0 ...

icon picture PDF Filetype PDF | Posted on 25 Jan 2023 | 2 years ago
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           CALCULUS TRIGONOMETRICDERIVATIVESANDINTEGRALS
                  TRIGONOMETRICDERIVATIVES
                    d                      0              d                        0            d               2       0
                    dx(sin(x)) = cos(x) · x              dx(cos(x)) = sin(x)·x                dx(tan(x)) = sec (x)·x
                    d                               0     d                             0       d                 2       0
                    dx(csc(x)) = csc(x)cot(x)·x         dx(sec(x)) = sec(x)tan(x)·x           dx(cot(x)) = csc (x)·x
                    d (sin1(x)) = p 1 2 · x0             d (cos1(x)) = p 1 2 ·x0             d (tan1(x)) =    1 2 · x0
                    dx                1x                dx                  1x               dx               1+x
                    d (csc1(x)) =  p1       · x0        d (sec1(x)) = p1       · x0          d (cot1(x)) =  1     · x0
                    dx                x x21             dx               x x21               dx                 1+x2
                  TRIGONOMETRICINTEGRALS
                   R sin(x)dx = cos(x)+C                                   R csc(x)dx =ln|csc(x)cot(x)|+C
                   R cos(x)dx =sin(x)+C                                     R sec(x)dx =ln|sec(x)+tan(x)|+C
                   R tan(x)dx =ln|sec(x)|+C                                 R cot(x)dx =ln|sin(x)|+C
                  POWERREDUCTIONFORMULAS                                       INVERSE TRIG INTEGRALS
                   R sinn(x)=1 sinn1(x)cos(x)+n1 R sinn2(x)dx               R sin1(x)dx = xsin1(x)+p1x2 +C
                                  n                      n
                   R cosn(x)=1 cosn1(x)sin(x)+n1 R cosn2(x)dx                R cos1(x)dx = xcos1(x)p1x2 +C
                                 n                      n
                   R tann(x)= 1 tann1(x)R tann2(x)dx                         R tan1(x)dx = xtan1(x) 1 ln(1+x2)+C
                                 n1                                                                           2
                   R     n          1     n1       R    n2                    R p dx          1x
                     cot (x)=n1cot          (x) cot       (x)dx                  a2x2 =sin      a +C
                   R    n         1            n2       n2 R    n2           R   dx      1    1x
                     sec (x)=n1 tan(x)sec         (x)+n1 sec        (x)dx       x2+a2 = a tan      a +C
                   R    n           1           n2       n2 R     n2         R pdx         1    1x
                     csc (x)=n1cot(x)csc          (x)+n1 csc        (x)dx      x x2a2 = a sec       a +C
                                                                                                         Tel: 
                  csusm.edu/stemsc                   XXX                            @csusm_stemcenter STEM SC (N): (760) 750-4101
                                                                                                         STEM SC (S): (760) 750-7324
         CALCULUS TRIGONOMETRICDERIVATIVESANDINTEGRALS
               STRATEGYFOREVALUATINGRsinm(x)cosn(x)dx
                 (a) If the power n of cosine is odd (n =2k +1), save one cosine factor and use cos2(x)=1sin2(x)to
                 express the rest of the factors in terms of sine:
                            Z sinm      n        Z    m      2k+1        Z    m       2   k
                                   (x)cos (x)dx =  sin (x)cos    (x)dx =   sin (x)(cos (x)) cos(x)dx
                                                                         Z    m          2   k
                                                                       = sin (x)(1sin (x)) cos(x)dx
                 Then solve by u-substitution and let u =sin(x).
                 (b) If the power m of sine is odd (m =2k + 1), save one sine factor and use sin2(x)=1 cos2(x)to
                 express the rest of the factors in terms of cosine:
                            Z sinm      n        Z    2k+1      n       Z     2   k   n
                                   (x)cos (x)dx =  sin    (x)cos (x)dx =  (sin (x)) cos (x)sin(x)dx
                                                                      =Z (1cos2(x))kcosn(x)sin(x)dx
                 Then solve by u-substitution and let u = cos(x).
                 (b) If both powers m and n are even, use the half-angle identities:
                                  sin2(x)=1(1cos(2x))              cos2(x)=1(1+cos(2x))
                                           2                                 2
                                                   R    m      n
               STRATEGYFOREVALUATING tan (x)sec (x)dx
                 (a) If the power n of secant is even (n =2k,k  2), save one sec2(x) factor and use sec2(x) = 1+tan2(x)
                 to express the rest of the factors in terms of tangent:
                      Z     m      n        Z    m      2k        Z    m       2 k1  2
                         tan (x)sec (x)dx =   tan (x)sec (x)dx =    tan (x)(sec )   sec (x)(x)dx
                                                                  Z    m           2   k1   2
                                                                = tan (x)(1+tan (x))      sec (x)(x)dx
                 Then solve by u-substitution and let u = tan(x).
                 (b) If the power m of tangent is odd (m =2k + 1), save one sec(x)tan(x) factor and use tan2(x)=
                 sec2(x) 1 to express the rest of the factors in terms of secant:
                      Z tanm       n        Z    2k+1      n       Z     2    k  n1
                             (x)sec (x)dx =   tan    (x)sec (x)dx =  (tan (x)) sec   (x)sec(x)tan(x)dx
                                                                 =Z (sec2(x)1)ksecn1(x)sec(x)tan(x)dx
                 Then solve by u-substitution and let u =sec(x).
                                                                                           Tel: 
               csusm.edu/stemsc              XXX                       @csusm_stemcenter
                                                                                           STEM SC (N): (760) 750-4101
                                                                                           STEM SC (S): (760) 750-7324
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...Calculus trigonometricderivativesandintegrals trigonometricderivatives d dx sin x cos tan sec csc cot p trigonometricintegrals r c ln powerreductionformulas inverse trig integrals sinn n xsin px cosn xcos tann xtan ncot ax a pdx xa tel csusm edu stemsc xxx stemcenter stem sc s strategyforevaluatingrsinm if the power of cosine is odd k save one factor and use to express rest factors in terms sine z sinm m then solve by u substitution let b kcosn both powers are even half angle identities strategyforevaluating secant tangent tanm ksecn...

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