Z tan x dx log cos x z cot x dx log sin x z sec 2 x

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Z tan x dx = - log | cos x | , Z cot x dx = log | sin x | , Z sec 2 x dx = tan x , Z cosec 2 x dx = - cot x , Z tan x sec x dx = sec x , Z cot x cosec x dx = - cosec x . [These integrals are included in the general form, but there is no need to use a substitution, as the results follow at once from § 119 and equation (5) of § 130 .] 3. Show that the integral of 1 / ( a + b cos x ), where a + b is positive, may be expressed in one or other of the forms 2 a 2 - b 2 arc tan ( t r a - b a + b ) , 1 b 2 - a 2 log b + a + t b - a b + a - t b - a , where t = tan 1 2 x , according as a 2 > b 2 or a 2 < b 2 . If a 2 = b 2 then the integral reduces to a constant multiple of that of sec 2 1 2 x or cosec 2 1 2 x , and its value may at once be written down. Deduce the forms of the integral when a + b is negative. 4. Show that if y is defined in terms of x by means of the equation ( a + b cos x )( a - b cos y ) = a 2 - b 2 , where a is positive and a 2 > b 2 , then as x varies from 0 to π one value of y also varies from 0 to π . Show also that sin x = a 2 - b 2 sin y a - b cos y , sin x a + b cos x dx dy = sin y a - b cos y ;
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[VI : 144] DERIVATIVES AND INTEGRALS 301 and deduce that if 0 < x < π then Z dx a + b cos x = 1 a 2 - b 2 arc cos a cos x + b a + b cos x . Show that this result agrees with that of Ex. 3. 5. Show how to integrate 1 / ( a + b cos x + c sin x ). [Express b cos x + c sin x in the form b 2 + c 2 cos( x - α ).] 6. Integrate ( a + b cos x + c sin x ) / ( α + β cos x + γ sin x ). [Determine λ , μ , ν so that a + b cos x + c sin x = λ + μ ( α + β cos x + γ sin x ) + ν ( - β sin x + γ cos x ) . Then the integral is μx + ν log | α + β cos x + γ sin x | + λ Z dx α + β cos x + γ sin x . ] 7. Integrate 1 / ( a cos 2 x +2 b cos x sin x + c sin 2 x ). [The subject of integration may be expressed in the form 1 / ( A + B cos 2 x + C sin 2 x ), where A = 1 2 ( a + c ), B = 1 2 ( a - c ), C = b : but the integral may be calculated more simply by putting tan x = t , when we obtain Z sec 2 x dx a + 2 b tan x + c tan 2 x = Z dt a + 2 bt + ct 2 . ] 144. Integrals involving arc sin x , arc tan x , and log x . The inte- grals of the inverse sine and tangent and of the logarithm can easily be calculated by integration by parts. Thus Z arc sin x dx = x arc sin x - Z x dx 1 - x 2 = x arc sin x + 1 - x 2 , Z arc tan x dx = x arc tan x - Z x dx 1 + x 2 = x arc tan x - 1 2 log(1 + x 2 ) , Z log x dx = x log x - Z dx = x (log x - 1) .
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[VI : 145] DERIVATIVES AND INTEGRALS 302 It is easy to see that if we can find the integral of y = f ( x ) then we can always find that of x = φ ( y ), where φ is the function inverse to f . For on making the substitution y = f ( x ) we obtain Z φ ( y ) dy = Z xf 0 ( x ) dx = xf ( x ) - Z f ( x ) dx. The reader should evaluate the integrals of arc sin y and arc tan y in this way. Integrals of the form Z P ( x, arc sin x ) dx, Z P ( x, log x ) dx, where P is a polynomial, can always be calculated. Take the first form, for example. We have to calculate a number of integrals of the type Z x m (arc sin x ) n dx . Making the substitution x = sin y , we obtain Z y n sin m y cos y dy , which can be found by the method of § 142 . In the case of the second form we have to calculate a number of integrals of the type Z x m (log x ) n dx . Integrating by parts we obtain Z x m (log x ) n dx = x m +1 (log x ) n m + 1 - n m + 1 Z x m (log x ) n - 1 dx,
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