7.2 Trigonometric Integrals

Transcription

1 7. Trigonometric Integrals The three identities sin x + cos x, cos x (cos x + ) and sin x ( cos x) can be used to integrate expressions involving powers of Sine and Cosine. The basic idea is to use an identity to put your integral in a form where one of the substituions u sin x or u cos x may be applied.. To compute sin 3 x dx, we first apply the identity sin x + cos x to get sin 3 x dx sin x( cos x) dx This is now set up perfectly for the substitution u cos x du sin x dx: sin 3 x dx ( u )( du) u du 3 u3 u + c 3 cos3 x cos x + c. The same trick works in the following: sin 3 x cos x dx sin x sin x cos x dx sin x( cos x) cos x dx sin x(cos x cos x) dx (u u )( du) (substitute u cos x) 3 u3 + 5 u5 + c 5 cos5 x 3 cos3 x + c 3. With an even power, a double-angle formula is more useful. sin x dx ( cos x) dx x + sin x + c General Strategies sin m x cos n x dx. If n k + is odd, replace cos k x ( sin x) k, then sin m x cos n x dx sin m x( sin x) k cos x dx Now substitute u sin x du cos x dx sin m x cos n x dx u m ( u ) k du This is a polynomial in u, which can be integrated after multiplying out.. If m is odd, repeat () with the roles of cos and sin reversed.. and. above are of this form.

2 3. If m k, n l are both even, write sin x ( cos x) and cos x ( + cos x) to obtain sin m x cos n x dx k+l ( cos x) k ( + cos x) l dx which yields integrals of powers of cos x. If these powers are odd, then we use the strategies above, if they are even we can repeat to find integrals in terms of cos x, etc. An alternative approach might utilise the identity sin x cos x sin x.. To compute sin x dx we have to use two identities: sin x ( cos x) and then cos x ( + cos x). sin x dx (sin x) dx ( cos x) dx cos x + cos x dx x sin x + cos x dx (integrate the bits you can... ) x sin x + + cos x dx (... use another identity) x sin x + sin x + c 3. If both powers of sine and cosine are odd, you may use either of the first two strategies: sin 5 x cos 3 x dx cos x sin 5 x( sin x) dx cos x(sin 5 x sin 7 x) dx u 5 u 7 du (substitute u sin x) 6 sin6 x 8 sin8 x + c Alternatively sin 5 x cos 3 x dx sin x( cos x) cos 3 x dx sin x(cos 3 x cos 5 x + cos 7 x) dx u 3 u 5 + u 7 ( du) (let u cos x) cos x + 3 cos6 x 8 cos8 x + c Since both these answers are correct, we must have discovered a (nasty) trig identity! Indeed by evaluating both at x we can quickly see that c c and the result is the evil-looking identity sin 6 x 3 sin 8 x 6 cos x + 8 cos 6 x 3 cos 8 x

3 3. This time we have a pair of even power. There are again a couple of approaches. or π π sin x cos x dx 8 sin x cos x dx π π ( cos x)( + cos x) dx + cos x cos x cos 3 x dx 8 ( π ) π 8 ( + cos x) + ( sin x) cos x dx ( π π ) 8 + sin x cos x dx ( π 8 + π) 6 sin3 x π 6 8 π π π (sin x cos x) cos x dx 8 sin x dx π ( cos x) dx π 6 6 sin x( + cos x) dx It is rarely obvious what strategy is going to be best, so don t be surprised if someone finds a faster way to do a problem. General Strategies for tan m x sec n x dx The next type of trigonometric integrals have the form tan m x sec n x dx. These are similar to, but trickier than, the sin m x cos n x dx integrals. The overall goal is exactly the same: use a trig identity to transform the integrand in such a way that an easy substitution will finish things off. The difficulty is that we have fewer easy identities involving tangent and secant to play with.. If n k is even take out a factor of sec x. Replace the remaining powers of secant using the identity sec x + tan!x, that is sec k x ( + tan x) k Our integral is now set up for the substitution u tan x. The idea is that we have extracted sec x du dx and left everythin else in terms of u tan x: tan m x sec n x dx tan m x( + tan x) k sec x dx u m ( + u ) k du which can be integrated after multiplying out.. If m k + is odd and n, save a factor of sec x tan x and use the identity tan x sec x to convert all the tangents to secants. We can then substitute u sec x, then du sec x tan x dx eliminates the saved expression. tan m x sec n x dx (sec x ) k sec n x sec x tan x dx (u ) k u n du which can be integrated after multiplying out. 3

4 In cases other than the above we may be stuck: we can sometimes appeal to the anti-derivatives tan x dx ln sec x + c, sec x dx ln sec x + tan x + c but these may be insufficient to solve the problem. Integration by parts or some other clever substitution may be necessary.. tan x sec x dx tan x( + tan x) sec x dx u ( + u ) du (substitute u tan x) 3 u3 + 5 u5 + c 3 tan3 x + 5 tan5 x + c. tan 5 x sec 3 x dx tan x sec x sec x tan x dx (sec x ) sec x sec x tan x dx (u ) u du (substitute u sec x) u 6 u + u du 7 u7 5 u5 + 3 u3 + c 7 sec7 x 5 sec5 x + 3 sec3 x + c 3. tan x sec x dx (sec x ) sec x dx sec 3 x sec x dx sec 3 x dx ln sec x + tan x sec 3 x dx can be attacked by parts: u sec x dv sec x dx } { du sec x tan x dx v tan x tan x sec x dx sec x tan x sec x tan x dx ln sec x + tan x tan x sec x dx (sec x tan x ln sec x + tan x ) + c Other Trigonometric Integrals Other trigonometric integrals will not tend to have general strategies: creativity is necessary! No other types of trigonometric integral are examinable in this class!

5 Suggested problems. Evaluate the integrals: (a) sin 3 x cos x dx (b) cos (x) dx. Evaluate the integrals: (a) sec x tan 3 x dx (b) π/3 sec 3 x tan 3 x dx 3. The region R is bounded by the graph of y tan x and the x-axis on the interval [, π/3]. The region R is bounded by the graph of y sec x and the x-axis on the interval [, π/6]. Which region has the greater area? 5