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(1)s=Int dx/(4+5cos x). By using t-substitution, i.e. t=tan(x/2) we get cos x=(1-t^2)/(1+t^2) and dx=2dt/(1+t^2). Substituting in s and simplifying, we get

s= 2 Int. dt/4(1+t^2)+5(1-t^2) = 2 Int. dt/(9-t^2).

Using standard result Int. dx/(a^2-x^2)=1/a*arctanh (x/a), we get

s=2/3*arctanh {1/3*tan(x/2)} +C1, which is correct as given in the book.

However, if we use further substitution of t=3 sin u, we get dt=3 cos u du and u=arcsin (t/3)

S=2/3*Int. cos u du/(1-sin^2 u)=2/3* Int. sec u du= 2/3*log[sec u+tan u)

=2/3*log[sec arcsin {1/3*tan(x/2)}+ tan arcsin {1/3*tan(x/2)}] +C2.

Though second procedure gives complicated result , when plotted both the results give same curve, but can we show analytically that the two results are same?

(2)Another similar question is integral of dx/(4-5 sin x). I have solved it using t-substitution with partial fractions and got correct answer as 1/3*[log{tan(x/2)-2}/{2tan(x/2)-1}] but can we solve it using only substitutions and without using partial fractions, as is required in my book.

(1) First, if you know what hyperbolic sine and cosine functions are, they are defined as linear combinations of e^x and e^(-x). In particular

sinh(M) = ( e^M - e^(-M))/2 or 1/2 *e^(M) - 1/2 *e^(-M)

and

cosh(M) =( e^M + e^(-M))/2 or 1/2 *e^(M) + 1/2 *e^(-M).

Following that, we can define tanh(M) as the quotient sinh(M)/cosh(M) which equals ( e^M - e^(-M))/( e^M + e^(-M)). As a consequence of that, with techniques in some solving of quadratic equations and the relation between exap and inverse of the exp function, tanh (M) is a disguised

for the function 1/2 * log ((1+M)/(1-M)). Hence using the first method , you really get

s= 2/3 * 1/2 * log ((1+M)/(1-M)) where M is 1/3 * t.

Second, sec(arcsin (t/3) ) and tan(arcsin (t/3 ) can both be expressed in terms of t with no reference to the trigonometric functions sin, cos, tan, cot, sec, csc or their inverses. This technique is fairly well-known in a course on trigonometry or Pre-Caclulus. For example if theta = arcsin (t/3), then we can also think of sin (theta) as t/3. So we can "draw" a right-angled triangle with angle theta and whose opposite is t and hypotenuse is 3.

| |\\ | \\ | \\ | \\ t | \\ 3 | \\ | \\ | \\ | \\ | \\ | theta\\ ------------\\ sqrt(9-t^2)

Using Pythagoras theorem, the adjacent side will be sqrt(9-t^2). So then this triangle will help us find all the trigonometric functions of theta in terms of the three sides of the triangle. In particular sec (arcsin (t/3) ) = sec (theta) = 3/sqrt(9-t^2) and tan (arcsin (t/3) ) =tan (theta) = t/sqrt(9-t^2) . So in essence, you second answer using trig substitution t=3 sin u and yielding an answer of 2/3*log[sec arcsin {1/3*tan(x/2)}+ tan arcsin {1/3*tan(x/2)}] is actually

2/3 * log [3/sqrt(9-t^2) + t/sqrt(9-t^2)]

= 2/3 * log [ (3+t)/sqrt(9-t^2)]

= 2/3 * log[3+t] - 2/3*log [sqrt(9-t^2)]

= 2/3 * log[3+t] - 1/3*log [(9-t^2)].

I feel like if I should go on, I am just doing a long of "manipulations" using either the tanh to log conversion or laws of logarithm, but it is now a simple challenge for you to see why the two results are the same analytically.

(2) If you do not like partial fraction in the integral of dx/(4-5 sin x) after that t-substitution (I call this the tan half -x formula), here's what you can do, try a t-substitution, then when you are at ntegral of dt/(2t^2 - 5t + 2), attempt a completing the square to make (2t^2 - 5t + 2) be equal to 2 (t-5/4)^2 - 9/8, then try a w-substitution with w = t-5/4 and you will be having integral of dw/ (2w^2 -9/8) Now bring out a factor of (-2) from the denominator, and you have a similar integral as in (1) and you could get a seemingly different answer but after understanding that many of these functions are the same but written as different "form", you know you're just calling hyperbolic sine when it's nothing but e^M and e^(-M) .

Calculus

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