sin xdx = cos x + c We also run into antiderivatives for tan x, cot x, sec x and csc x in the section on Log integrals. They are: cos ax sec ax a

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1 Trig Integrals We already know antiderivatives for sin x, cos x, sec x tan x, csc x, sec x and csc x cot x. They are cos xdx = sin x sin xdx = cos x sec x tan xdx = sec x csc xdx = cot x sec xdx = tan x csc x cot xdx = csc x We also rn into antiderivatives for tan x, cot x, sec x and csc x in the section on Log integrals. They are: tan xdx = ln cos x = ln sec x cot xdx = ln sin x sec xdx = ln sec x + tan x csc xdx = ln csc x ot x It is easy to generalize to antiderivatives of sin ax, cos ax, sec ax tan ax, csc ax, sec ax, csc ax cot ax tan ax and cot ax where a is a non-zero constant. Indeed sin ax cos ax sec ax cos axdx =, sin axdx =, sec ax tan axdx =, etc a a a We now wish to address antiderivatives of polynomials and rational fnctions involving trigonometric fnctions; i.e. antiderivatives of fnctions like sin x, sin x cos x, tan 3 cos x ax,, etc. We will also address os x antiderivatives that may be traced back to inverse trig fnctions. Antiderivatives of sqares of trig fnctions To find sin xdx, cos xdx, tan xdx and cot xdx note that sin x = cos x, cos x = + cos x, tan x = sec x and cot x = csc x It follows that: ( sin xdx = cos x) dx = x sin x, ( cos xdx = + cos x) dx = x + sin x, (sec tan xdx = x ) dx = tan x x, (csc cot xdx = x ) = cot x x Antiderivatives of higher powers of trig fnctions For third and higher powers of trig fnctions, we se the following redction formlas which yo may have rn into already:

2 . If n 0 then cos n xdx = cosn x sin x + n cos n xdx. n n For example, to find cos xdx we apply the redction formla with n = to get We then apply it to cos xdx = cos x sin x cos 3 xdx with n = 3 to get + cos 3 xdx cos xdx = cos x sin x + [ cos x sin x + ] cos xdx 3 3 We do not need a redction formla to evalate cos xdx. The final answer is cos xdx = cos x sin x + cos x sin x + 8 sin x. If n 0 then sin n xdx = sinn x cos x + n sin n xdx. n n For example, to find sin xdx we apply the redction formla with n = to get We know that sin xdx = sin3 x cos x + 3 sin xdx sin xdx = x sin x. Therefore the final answer is sin xdx = sin3 x cos x + 3 ( x sin x) = sin3 x cos x + 3x 8 3 sin x 6 3. If n then tan n xdx = tann x tan n xdx. n For example, to find tan 3 xdx we apply the redction formla with n = 3 to get tan 3 xdx = tan x tan xdx = tan x ln sec x. If n then sec n xdx = secn x tan x + n sec n xdx. n n For example, to find sec 6 xdx we apply the redction formla with n = 3 to get We then apply it to sec xdx to get sec 6 xdx = sec tan x + sec xdx sec 6 xdx = sec tan x + [ sec x tan x ] sec xdx

3 We already know sec xdx. Therefore the final answer is sec 6 xdx = sec tan x + sec x tan x + 8 tan x Exercise. Determine each antiderivative: (a) tan xdx sin xdx cos 6 xdx (d) sec xdx (e) tan 3 xdx (f) sec xdx. Use the redction formla for sec n xdx to show that sec 3 xdx = [sec x tan x + ln sec x + tan x ]. (Expression in cos x) sin xdx (cos Examples of sch integrals are x + 3 ) ( ) sin xdx and cos x 3 cos x + sin xdx. They be redced to more familiar integrals by applying the sbstittion = cos x. ( ) Example To determine cos x 3 cos x + sin xdx, let = cos x. Then the integral is transformed into ( ) 3 + d. which we can easily handle to get ln = ln cos x os 3 x + cos x. Example 3 To determine cos x sin 3 xdx, first write it as cos x ( sin x ) sin xdx = cos x ( cos x ) sin xdx Now a sbstittion = cos x gives cos x ( cos x ) ( sin xdx = ) d = cos x cos3 x 3 (Expression in sin x) cos xdx These may be redced to more familiar integrals by making a sbstittion = sin x (Expression in tan x) sec xdx These may be redced to more familiar integrals by making a sbstittion = tan x. 3

4 (tan Example To determine 3 x + ) sec xdx, let = tan x, This transforms the integral into ( 3 + ) d which we can easily handle. We get (tan 3 x + ) ( sec xdx = 3 + ) d = + = tan x + tan x. (Expression in sec x) sec tan xdx These may be redced to more familiar integrals by making a sbstittion = sec x. Example To determine tan x sec 3 xdx, first write it as (tan x sec x ) sec x tan xdx = [(sec x ) sec x] sec x tan xdx. Now set = sec x. Then tan x sec 3 xdx = ( ) ( d = 6 + ) d (Expression in csc x) csc x cot xdx are handled in a similar way. Exercise 6 Determine each integral cos x (a) sin 3 x dx (d) sec x tan xdx (e) = = 7 sec7 x sec x + 3 sec3 x (Expression in cot x) csc xdx and sin x cos xdx sin x cos xdx (f) sin 3 x cos 3 xdx sec x tan xdx (g) cos 3 x sin 3 x xdx (tan (h) x sec x ) sin xdx (cot (i) 3 x + ) csc xdx Integrals Involving Prodcts of sin ax and cos bx Examples of sch integrals are sin x cos 3xdx, sin x sin 7xdx, cos x cos 3xdx. We redce them to familiar integrals by sing identities that convert them into integrals of sms involving sin px or cos qx. For example, since sin x cos 3x = [sin (x + 3x) + sin (x 3x)] = [sin 7x + sin x] we replace sin x cos 3xdx with [sin 7x + sin x] dx which we know how to handle. The identities to se are: sin A cos B = [sin(a + B) + sin(a B)] cos A cos B = [cos(a B) os(a + B)] sin A sin B = [cos(a B) cos(a + B)]

5 Exercise 7 Determine each integral (a) sin x cos xdx cos 3x cos xdx sin 6x sin 7xdx (d) cos πx sin xdx (e) sin 3πx sin xdx (f) cos ax cos bxdx Integrals Involving x Since the derivative of arcsin x is x, an integrand with the term x may be the reslt of taking the derivative of an expression involving arcsin x. This sggests a sbstittion = arcsin x, which is eqivalent to x = sin. One reslt of the sbstittion x = sin is that x is replaced by sin = cos = cos, ths getting rid of the sqare root. Example 8 To determine integral becomes x x = = x dx dx, set x = sin. Then x d sin sin sin cos d = cos d = cos ( cos ) d = sin = cos, and dx = cos d. The sin d Since sin = sin cos and cos = sin = x, we may write the above reslt as x = x arcsin x (x ) x = (arcsin x x ) x x Example 9 Consider / x dx. Let x = sin. Then dx d x =, = π 6 and when x =, = π. Therefore = cos, and dx = cos d. When / x x dx = = π/ π/ sin sin cos d = π/ cot d ( csc ) [ ] π/ d = cot + = 3 3 π. Integrals involving ( + x ) Since the derivative of arctan x is, an integrand with the term may be the reslt of differentiating an expression related to arctan x. This sggests a sbstittion = arctan x, or x = tan. One + x + x conseqence is that + x is replaced by a single term, namely + tan = sec. dx. Example 0 To determine x dx, set x = tan. Differentiating gives + x d = sec. Therefore dx = sec d and the integral becomes sec tan + tan d = sec cos tan sec d = sin d = sin. Since x = tan, sin = x, hence + x x + x dx = + x. x

6 Integrals involving x Since sec = tan, a sbstittion x = sec x may simplify an integral involving x by getting rid of the sqare root. More precisely, if we let x = sec then x = sec = tan = tan and the sqare root is gone. (Note that x is different from x.). Example Consider x dx. Let x = sec. Then dx = sec tan d and the integral x becomes sec tan sec tan d = d =. To express this in terms of x, note that x = sec implies that cos = x, therefore = arccos x, and so x x dx = arccos x. Example To determine becomes sec (sec tan ) d = tan Yo already showed that x dx. Let x = sec. Then dx = sec tan d and the integral x sec 3 d. sec 3 dx = [sec tan + ln sec + tan ]. Since sec = x, it follows from the identity tan = sec that tan = x, therefore x x dx = [ x x + ln x + ] x Exercise 3. Determine each integral (a) x x dx x + x dx x x dx x (d) dx (e) x ( + x dx (f) ) (x + ) (g) dx (h) ( x 3/ ) dx x (i) x dx x x + (j) (x dx (k) dx (l) + ) ( + x 3/ ) dx + x (m) x dx x x dx x dx 3/ dx. Use the sbstittion x = cos to show that ( + / x x3 ( x) dx = + ) x x 6

7 3. Use the sbstittion x = tan to show that ( ( x x + ) + ) / ( x + ) 3/ dx = 3. Use the sbstittion x = sin to show that / / ( ) / x dx = π 3 x +. Use the sbstittion = x x to show that 0 + x dx = π 8 x (sec 6. Show that the sbstittion x = tan transforms dx into 3 sec ) d then deter- + x mine x dx + x 0 7