Math F15 Rahman

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1 Math - 9 F5 Rahman Week3 7.3 Hyperbolic Functions Hyperbolic functions are similar to trigonometric functions, and have the following definitions: sinh x = (ex e x ) cosh x = (ex + e x ) tanh x = sinh x cosh x cschx = / sinh x sechx = / cosh x coth x = / tanh x It s also useful to know what they look like To remember what they look like, just use the definitions and recall what the exponential functions look like and take the average. If you re confused as to what I m talking about make sure to ask me to explain it. They are subject to the following identities: sinh( x) = sinh x cosh( x) = cosh x cosh x sinh x = tanh x = sech x sinh(x + y) = sinh x cosh y + cosh x sinh y cosh(x + y) = cosh x cosh y + sinh x sinh y We can prove some of these things, so we may get a better understanding of the identities. Proofs are important, even for engineers!

2 Theorem. cosh x sinh x =. Proof. We go straight to the definition, cosh x sinh x = [ ] [ ] (ex + e x ) (ex e x ) = 4 (ex + + e x ) 4 (ex + e x ) = Theorem. tanh x = sech x Proof. Here we simply divide the entire equation by cosh x, [ cosh x sinh x = ] cosh x tanh x = sech x. The other identities are proved similar to this one. If you have time, you should try to prove the other identities by yourselves. Even though they wont appear on exams they will help you get a better understanding of the concepts. Here is a nice proof of one of the most important trigonometric identities, and all other identities can be very easily derived through these identities in a similar fashion to the above theorem. Theorem 3. sin θ + cos θ =. Proof. Consider a right triangle and one non-right angle θ. Let the side opposite to θ be of length x, the side adjacent to θ be of length y, and the hypotenuse z. Then, sin θ = x/z and cos θ = y/z, and by the Pythagorean theorem x + y = z, then sin θ + cos θ = x z + y z = x + y z = z z =. It is important to know the derivatives of hyperbolic functions as well, (sinh x) = cosh x (cosh x) = sinh x (tanh x) = sech x (cschx) = cschx coth x (sechx) = sechx tanh x (coth x) = csch x These can all be derived very easily from the definitions. Ex: (cosh x) = x (sinh( x)).

3 8. Integration Review Here we do some standard book problems from the section. 4 I = π/3 π/4 dx/(cos x tan x). Solution: I = π/3 π/4 dx/ cos x = π/3 π/4 sec xdx = ln sec x+tan x π/3 ln + 3 ln +. 8 I = e y dy/ y Solution: Let u = y, then I = u du = u / ln = y / ln. 4 I = xdx/( + x 3 ). Solution: Let u = x 3/ du = 3 xdx/, then I = 3 du/( + u ) = 3 tan u = 3 tan (x 3/ ) + C. π/4 = 8. Integration by parts The modern notion of integration by parts comes from a beautiful theory of integrals by Riemann and Stieltjes in 894, soon after which Stieltjes passed away. The idea is we can integrate over certain functions instead of just over x. We can think of it as a generalization of u-sub. To derive it, consider the product rule, d dx [f(x)g(x)] = f(x)g (x) + g(x)f (x) d[f(x)g(x)] = f(x)g (x)dx + g(x)f (x)dx d[f(x)g(x)] = f(x)g(x) = f(x)g (x)dx + g(x)f (x)dx f(x)g (x)dx = f(x)g(x) g(x)f (x)dx. This can be written in the form, which we will use from now on () udv = uv vdu. () I = x sin xdx. Solution: Let u = x du = dx and dv = sin x v = cos x. Then, I = x cos x + cos xdx = x cos x + sin x + C.

4 We see here that we generally choose the easiest thing to integrate as dv. We can use ILATE: InverseLogsAlgebraicTrigonometricExponential, to help determine which is easier to integrate. Things get easier to integrate as we go to the right, for example, Exponentials are easier to integrate than Trigonometric functions. But this doesn t always work! So, only use it as a guide, not a rule of thumb. () I = ln xdx. Solution: Let u = ln x du = xdx and dv = dx v = x. Then, I = x ln x x dx = x ln x x + C. x (3) I = t e t dt. Solution: Let u = t du = tdt and dv = e t dt v = e t. Then, I = t e t te t dt. Notice, we need to integrate by parts again for the second integral I = te t dt. Let u = t du = dt and dv = e t dt v = e t. Then I = te t e t dt = te t e t. Plugging this back into I gives, I = t e t te t + e t + C. It may be appealing to do this sort of problem using tabular integration, however you should avoid using this method. If you make a mistake using this method, you will lose a majority of the points. You are better off doing integration by parts twice. (4) I = e x sin xdx. Solution: Let u = e x dx du = e t dt and dv = sin x v = cos x. Then, I = e x cos x + e x cos xdx. We must do another integration by parts on the second integral. Let u = e x du = e x dx and dv = cos x v = sin x. Then, I = e x sin x e x sin xdx. Plugging this into I gives, I = e x sin x e x cos x e x sin xdx. Now, we add both sides by e x sin xdx, to get e x sin xdx = e x sin x e x cos x e x sin xdx = (ex sin x e x cos x)+c.

5 Notice, for this problem it didn t matter if you chose e x or sin x and cos x as your u or dv. Try this problem the other way around to convince yourself that it works both ways. And as usual, if you re confused about what I m talking about, please make sure to ask me. It s better to get questions answered early on before you re bombarded with new material. (5) I = tan xdx. Solution: Let u = tan x I = x tan x dx +x and dv = dx v = x. Then, xdx + x. The second integral is our usual u-sub integral where u = + x du = xdx. Then, I = du u = ln u Plugging this back into I gives, I = x tan x = ln. ln u = π 4 ln (6) This next example is a test of our abilities to think abstractly. You wont see this sort of thing on the exam, but you ll see things on the exam that use many of the tricks we will use on this example. Find a reduction formula for I = sin n xdx. Solution: Let u = sin n x du = (n ) sin n x cos xdx and dv = sin xdx v = cos x. Then, sin n xdx = cos x sin n x + (n ) sin n x cos xdx = cos x sin n x + (n ) sin n x( sin x)dx = cos x sin n x + (n ) sin n x (n ) sin n xdx n sin n xdx = cos x sin n x + (n ) sin n xdx sin n xdx = n cos x sinn x + n n sin n xdx.

6 8.3 Trigonometric Integration Lets look at a few examples first and then we ll develop a general strategy. Sines and Cosines. () Consider cos 3 xdx. Solution: We recall the identity: cos = sin x. Lets see if we can use this to simplify the problem. cos 3 xdx = cos x[ sin x]dx = cos xdx sin x cos xdx. Now, the first integral is easy and the second integral we solve via u-sub where u = sin x du = cos xdx. cos 3 xdx = sin x u du = sin x 3 u3 + C = sin x 3 sin3 x + C () sin 5 x cos xdx. Solution: Lets use the same strategy as above, except this time on sin x. sin 5 x cos xdx = (sin x) sin x cos xdx = ( cos x) cos x sin xdx. We can go straight to u-sub with u = cos x du = sin x, sin 5 x cos xdx = ( u ) u du = 3 u3 + 5 u5 7 u7 +C = 3 cos3 x+ 5 cos5 x 7 cos7 x+c (3) π sin xdx Solution: For this problem if we used the identity we used for the past two problems we would be going in circles, so we use another identity - the double angle formula: cos x = sin x, π sin xdx = π ( cos x)dx = (4) sin 4 xdx Solution: This is similar to the above problem, [ ( x )] π sin x = π [ sin 4 xdx = = 4 ] ( cos x) dx = 4 [ cos x + ( + cos 4x) ] dx = 4 ( cos x + cos x)dx ( 3 x sin x + sin 4x 8 ) + C

7 Strategies for sin m x cos n xdx. () () If the power of the cosine term is odd (i.e. n = k + ), save one cosine factor and use cos x = sin x, sin m x cos k+ dx = = sin m x(cos x) k cos xdx sin m x( sin x) k cos xdx. Then substitute u = sin x du = cos x. () If the power of the sine term is odd (i.e. m = k + ), save one sine factor and use sin x = cos x, (3) sin k+ cos n xdx = (sin x) k cos n xdx = ( cos x) k cos n x sin xdx. Then substitute u = cos x du = sin x. (3) If the powers of both sine and cosine are even, use the double-angle formulas: sin x = ( cos x) cos x = ( + cos x) sin x cos x = sin x. Tangents and Secants. () tan 6 x sec 4 xdx. Solution: We recall the identity sec x = + tan x, and see where this takes us tan 6 x sec 4 xdx = tan 6 x( + tan x) sec xdx. Then we substitute u = tan x du = sec xdx, then tan 6 x sec 4 xdx = u 6 (+u )du = 7 u7 + 9 u9 +C = 7 tan7 x+ 9 tan9 x+c

8 () tan 5 θ sec 7 θdθ. Solution: Here lets try using the other identity: tan x = sec x, tan 5 θ sec 7 θdθ = tan 4 θ sec 6 θ sec θ tan θdθ = (sec θ ) sec 6 θ sec θ tan θdθ. We employ the u-sub u = sec θ du = sec θ tan θdθ, tan 5 θ sec 7 θdθ = (u ) u 6 du = u 9 u9 + 7 u7 +C = sec x 9 sec9 x+ 7 sec7 x+c Strategies for tan m x sec n xdx. () If the power of the secant term is even (i.e. n = k, k ), save a factor of sec x and use sec x = + tan x, (4) (5) tan m x sec k xdx = = tan m x(sec x) k sec xdx tan m x( + tan x) k sec xdx. Then substitute u = tan x du = sec xdx. () If the power of the tangent term is odd (i.e. m = k + ), save a factor of sec x tan x and use tan x = sec x, tan k+ x sec n xdx = (tan x) k sec n x sec x tan xdx = (sec x ) k sec n x sec x tan xdx. Then substitute u = sec x du = sec x tan xdx Useful Integrals. These integrals are also pretty easy to derive if you forget them, (6) tan xdx = ln cos x + C = ln sec x + C. (7) sec xdx = ln sec x + tan x + C. () tan 3 xdx. Solution: We use the identity tan x = sec x, tan 3 xdx = tan x(sec x )dx = tan x ln sec x + C.

9 () sec 3 xdx. Solution: We integrate by parts with u = sec x du = sec x tan xdx and dv = sec x v = tan x, then sec 3 xdx = sec x tan x = sec x tan x sec x tan xdx = sec x tan x sec x(sec x )dx sec 3 xdx + sec xdx = sec x tan x sec 3 xdx + ln sec x + tan x sec 3 xdx = [sec x tan x + ln sec x + tan x ] + C.