Задача 33296 ...

leydi09

5 февраля 2019 г. в 18:09

13. ∫ sin³ 4x dx
14. ∫ cos⁴ x/5 dx
15. ∫ sin⁴ x/3 cos² x/3 dx
17. ∫ tg³ 4x dx
18. ∫ dx / 3cos 4x – 2 sin 4x + 1

sova

5 февраля 2019 г. в 22:11

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13.
sin³4x=sin²4x·sin4x=(1–cos²4x)·sin4x;

∫ sin³4xdx= ∫ (1–cos²4x)·sin4xdx= ∫ sin4xdx– ∫ cos²4x·sin4xdx=

=(1/4)·(–cos4x)– ∫ cos²4xd(cos4x)/(–4)=

=(1/4)·(–cos4x)+(1/4)·(cos³4x)/3+C=
=(–1/4)·cos4x+(1/12)cos³4x +C
14.
cos⁴(x/5)=(cos²(x/5))²=((1+cos(2x/5))/2)²=

=(1/4)+(1/2)cos(2x/5)+(1/4)cos²(2x/5)=

=(1/4)+(1/2)cos(2x/5)+(1/4)·(1+cos(4x/5))/2=

(1/4)+(1/8)+(1/2)cos(2x/5)+(1/8)cos(4x/5)

∫cos⁴(x/5)dx= ∫ ((3/8)+(1/2)cos(2x/5)+(1/8)cos(4x/5))dx=

=(3/8)x+(5/4)sin(2x/5)+(5/32)sin(4x/5)+C

15.
sin⁴(x/3)·cos²(x/3)=(sin²(x/3))²·cos²(x/3)
применяем формулы понижения степени:
(1–cos(2x/3))²·(1+cos(2x/3)=

=(1–cos²(2x/3))·(1–cos(2x/3))=sin²(2x/3)·(1–cos(2x/3)

∫ sin⁴(x/3)·cos²(x/3)dx= ∫ sin²(2x/3)·(1–cos(2x/3)dx=

= ∫sin²(2x/3)dx – ∫ sin²(2x/3)cos(2x/3)dx=

= ∫ (1–cos(4x/3))dx/2 –(3/2) ∫ sin²(2x/3)d(sin(2x/3))=

=(1/2)x – (3/8)·sin(4x/3)–(3/2)·(sin³(2x/3))/3 +C=

=(1/2)x – (3/8)·sin(4x/3)–(1/2)·(sin³(2x/3)) +C

17.
ctg³4x=ctg4x·(ctg²4x)=ctg4x·((1/sin²x)–1)

d(ctg4x)=(ctg4x)`dx=(–1/sin<sup>2</sup>4x)·(4x)`=–4dx/sin²4x


∫ ctg³4xdx= ∫ ctg4x·dx(1/sin²x)– ∫ ctg4x=

=(–1/4) ∫ ctg4x d(ctg4x)– ∫ cos4xdx/sin4x=

=(–1/4)(ctg²(4x))/2–(1/4) ∫ d(sin4x)/sin4x=

(–1/8)ctg²(4x)–(1/4)ln|sin4x|+С

18.
3cos4x–2sin4x+1=3·(cos²2x–sin²2x)–2·2sin2xcos2x+sin²2x+cos²2x=

=4cos²2x–4sin2x·cos2x–2sin²2x= cos²2x·(4–4tg2x–2tg²2x)

tg2x=t
–2t²–4t+4=–2(t²+2t+1)+6=–2(t+1)²+6=6–2(t+1)²
d(tg2x)=(tg2x)`dx=(2x)`dx/cos²2x
dx/cos²(2x) = dt/2

получим табличный интеграл
(1/4)∫dt/(3–(t+1)²)=(1/4) ·(1/(2√3))ln | (√3+tg2x+1)/(√3–tg2x–1)| + C