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            Integrals of Trigonometric Functions 
            Starter
            1.     (Review of last lesson)  
                   A curve has gradient  4  and passes through the point (2, 7).  Susan says the equation of 
                                        3x       4                     4
                   the curve could be either y = 3 ln x + 6.08 or y = 3 ln3x + 4.61.  Tom said she is 
                   wrong because the equation must be unique.  Who is correct and why?
            2.     Given that d(sin x) = cos x, d(cos x) = − sin x and d(tan x) = sec2 x, find:
                                 dx                 dx                       dx
                   (a)      ∫ k sin xdx        (b)    ∫k cosxdx            (c)    ∫k sec2 xdx.
            Notes
            Always differentiate back mentally to check your answer is correct…and include the constant of 
            integration.
                                                ∫sin xdx = −cosx +c
                                                ∫cosxdx = sin x +c
                                               ∫sec2 xdx = tan x + c
            N.B.   When using trigonometric in calculus, angles must be in radians.
            E.g. 1 Find:  (a)    ∫ 1 cosxdx                  (b)      ∫ 6 sin x − 7 sec2 xdx
                                    7
                   Working:      (a)     ∫ 1 cosxdx = 1 sin x + c
                                           7              7
                          π
            E.g. 2 Find ∫ 6 5cos xdx.
                         0
                                     π
            E.g. 3 Find k such that ∫ 3 7sin xdx = 1.  Give your answer to 3 s.f.
                                     k               3
                                                                             Video:  Integrating sin/cos/tan
                                                                              Solutions to Starter and E.g.s 
            Exercise
            p193 9B Qu 1idef, 3ifg, 5, 6, 9, 12 (not Qu 10)
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           Qu 12(a)     You need to use your calculator to solve ex − 5sin x = 0.
                                                        Video:       Solving equations with Classwiz 
           	      	     Make sure your calculator is in radians.

                              ex − 5sin x = 0
           	      	     Enter                on your Classwiz using

           	      	     	      Alpha >> ) to enter “x”

           	      	     	      ALPHA >> CALC to enter “=”

           	      	     Then press SHIFT >> CALC to SOLVE

           	      	     For some reason your calculator throws up an old value for x that is in its memory.

           	      	     Press “=” to get the actual value you want which is 0.263265.

           Summary
             ∫ Aekxdx = A∫ekxdx
           ∫ P dx = P ∫ 1dx = P ln|x|+c
             Qx        Q x          Q
           ∫k sin xdx = −k cosx +c
           ∫k cosxdx = k sin x +c
           ∫k sec2 xdx = k tan x + c
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