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The three results obtained are equivalent since /l tan;\4tM 1 \ 2 ) t+tan x/2 tanjr/4+ti , l - t,... 3 Integrals of squared functions Examples 5. Integrate Ix = cos2 x dx, I2 = sin2 x ax. Solution. We give two methods of solution. 54 PROBLEMS AND METHODS IN ANALYSIS (i) Writing cos2 x = — (1 +cos 2x), we obtain at once Ι± = — (1 +cos 2x) ax, = y ( x + y s i n 2 x j + C. Similarly, we write sin2 x = —- (1 —cos 2x) and obtain M (1 — cos 2x) dx, = 2"(*-■2 s i n 2 « l + C'. CO A+^2 = (cos2 x+sin 2 x) dx, ^Jdx, = X, (cos2 x— sin2 x) dx, I±—12 = = cos2xdx i ' 1 · 2JC.

48 PROBLEMS AND METHODS IN ANALYSIS Thus in terms of x, 7=tanh-i{^2*; + 1>} + C, or as a logarithmic function, /=4log x + yj(2x2 + l) x-J(2x2 + l) EXERCISE + C. 21 Obtain the following integrals: (8*+3) 1. V(4* 2 +3*+l) 3. V(2*-;*:2) * 5. d*. 4. j (* + 3) djc. C _(x± J V(l-4^ 2 ) 8. 9. f V(l-4* s )d*. 10. 11. J Λ/(5+4Λ:-Λ: 2 ) r *-5 ax. 19. 21. 23. 12. d* dx d* r >0. V(2*r-*2),' x V ( l - 2 * - 3 ; t 2 ) d*. 6JC + 5 V(6+X-A:2) f jc + l J V(8+2JC-X 2 ) dx j V(* -*+m) 2 18. : x+3 d*. V(*2+2*) j y/(x -ax) 2 dx, 3Λ:+2 V(^2-4JC+5) 20.

F 24. f 25. Ϊ x2^/(7-2x)dx. 26. f ^ · J"V(1 + V*><*- ^—- V(*+i)+V(*+i) 29. f_ ί d *. 30. 2 fii»cffo/w o//Ae / W e ,y(ßx2+öx+c) We consider first the two following integrals dx / and vV-*2) where | x | < a, dx it a*, n » w n e r e x 2 + fc =- 0. f (a) dx 2 2 J V(« -* ) Put x = a sin Θ, then dx = a cos 0 d0, and T- f a cos Θ do a2 ß2sin2e J V( 1*1 J Λ dö . X i—- arc sin —. |a| a ) dx INDEFINITE INTEGRALS 35 X Thus for a > 0, / = arc sin —h C, and for # < 0, 1= — arc sin —h C, X — HYO ein —a Lr1 X = arc sin Therefore, C 1 dx 2 2 x J V(« -* ) QYC Clt!

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