найти интеграл - страница 18
Интеграл от 3pi/8 до pi/8
12 sin(pi/8 -x) cos (pi/8 - x) dx
Решение: π/8 π/8 3 π/8
∫12 sin(π/8-x)·cos(π/8-x)dx= ∫ 6sin(π/4-2x)dx=6∫sin(π/4-2x)dx=
3π/8 3π/8 π/8
3π/8
=3cos(π/4-2x)l= =3(cos(π/4-6π/8)-cos(π/4-2π/8)=3(cos(π/4-3π/4)-
π/8
-cos0=3 (0-1)=3
интеграл x корень x^2-5 dx
Решение: $$ \int x\sqrt{x^2-5}\, dx=[\, x=\frac{\sqrt5}{cost},\; dx=\frac{\sqrt5sint}{cos^2t}dt\, ]=\\\\=\int \frac{\sqrt5}{cost}\cdot \sqrt{\frac{5}{cos^2t}-5}\cdot \frac{\sqrt5sint}{cos^2t}\, dt=\int \frac{\sqrt5}{cost}\cdot \sqrt{5tg^2t}\cdot \frac{\sqrt5sint}{cos^2t}\, dt=\\\\=5\sqrt5\cdot \int \frac{tg^2t\cdot sint}{cos^3t} dt=5\sqrt5\cdot \int \frac{sin^3t}{cos^5t} dt=\\\\=5\sqrt5\int tg^3t\cdot \frac{dt}{cos^2t}=5\sqrt5\int tg^3t\cdot d(tgt)=5\sqrt5\cdot \frac{tg^4t}{4}+C= \\ =\frac{5\sqrt5}{4}\cdot tg^4(arccos\frac{\sqrt5}{x})+C $$Найдите интеграл \( \int\limits^7_0 { \sqrt{ 49- x^{2} }} \,x dx \)
Решение: 7
∫(√(49 -x²)dx)
0
замена: x =7sint,0=7sint⇒ t₁=0 ;7=7sint ⇒t₂=π/2 ;dx =7costdt ;t₂=π/2 ||
π/2 π/2 π/2
∫(7cost*7costdt) =49∫(1+cos2t)/2 dt =(49/2)(t +(1/2)sin2t) |
0. 0.0
= (49/2)*(π/2 +(1/2)sin2*π/2 - 0 -(1/2)sin2*0) =
=(49/2)*(π/2 +(1/2)sinπ - 0 -(1/2)sin0) =(49/2)*(π/2 +0 - 0 -0)
= 49π/4.
Интеграл
\( \int\limits^0_{-1} { \sqrt{1-x^{2} } } \, dx = \frac{ \pi}{4} \)
Почему так и каким образом это вычисляется?
Решение: Integral [-1;0] dx корень(1-x^2) =.
{x=sin(t);dx=cos(t)dt}
. = integral [3pi/2;2pi] dt cos^2(t) =
= integral [3pi/2;2pi] dt (cos(2t)+1)/2 =
= integral [3pi/2;2pi] dt cos(2t)/2 + integral [3pi/2;2pi] dt 1/2 =
= sin(2t)/4 [3pi/2;2pi] + t/2 [3pi/2;2pi]=
= 0 + pi/4 = pi/4
$$ \int\sqrt{1-x^2}dx\Rightarrow \left|\begin{array}{ccc}x=sint\\dx=cosdt\end{array}\right|\Rightarrow\int\sqrt{1-sin^2t}\cdot costdt\\\\\\=\int\sqrt{cos^2t}\cdot costdt=\int cost\cdot costdt=\int cos^2tdt=(*)\\-\\cos2t=2cos^2t-1\to2cos^2t=cos2t+1\to cos^2t=\frac{1}{2}(cos2t+1)\\-\\\\(*)=\int\left[\frac{1}{2}(cos2t+1)\right]dt=\frac{1}{2}\int(cos2t+1)dt=\frac{1}{2}\int cos2tdt+\frac{1}{2}\int dt \\ =\frac{1}{2}sin2t\cdot\frac{1}{2}+\frac{1}{2}t=\frac{1}{4}sin2t+\frac{1}{2}t=\frac{1}{4}\cdot2sintcost+\frac{1}{2}t=\frac{1}{2}sintcost+\frac{1}{2}t\\\\=\frac{1}{2}(sintcost+t)=(*)\\-\\x=sint\Rightarrow t=arcsinx\\\\sin^2t+cos^2t=1\to cos^2t=1-sin^2t\to cost=\sqrt{1-sin^2t}\\\to cost=\sqrt{1-x^2}\\-\\\\(*)=\frac{1}{2}(x\sqrt{1-x^2}+arcsinx) $$
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$$ \int\limits_{-1}^0\sqrt{1-x^2}dx=\left[\frac{1}{2}(x\sqrt{1-x^2}+arcsinx)\right]^0_{-1}\\\\=\frac{1}{2}(0\sqrt{1-0^2}+arcsin0)-\frac{1}{2}(-1\sqrt{1-(-1)^2}+arcsin(-1))\\\\=\frac{1}{2}\cdot0-\frac{1}{2}(-1\sqrt{1-1}-\frac{\pi}{2})=-\frac{1}{2}\cdot(-\frac{\pi}{2})=\frac{\pi}{4} $$
Интеграл cos^2x*dx/sin^4x
интеграл dx/cos^4x*sin^2x
Решение: $$ \int \frac{cos^2xdx}{sin^4x}=\int \frac{cos^2x}{sin^2x}\cdot \frac{dx}{sin^2x}=\int ctg^2x\cdot \frac{dx}{sin^2x}=[t=ctgx,\; dt=-\frac{dx}{sin^2x}]=\\\\=-\int t^2\cdot dt=-\frac{t^3}{3}+C=-\frac{ctg^3x}{3}+C\\\\\\\int \frac{dx}{cos^4x\cdot sin^2x}=\int \frac{\frac{dx}{cos^6x}}{\frac{cos^4xsin^2x}{cos^6x}}=\int \frac{(\frac{1}{cos^2x})^2\cdot \frac{dx}{cos^2x}}{tg^2x}=\\\\=[t=tgx,dt=\frac{dx}{cos^2x},1+tg^2x=\frac{1}{cos^2x}]=\int \frac{(1+t^2)^2dt}{t^2}=\int \frac{1+2t^2+t^4}{t^2}dt= \\ =\int (t^{-2}+2+t^2)dt=\frac{t^{-1}}{-1}+2t+\frac{t^3}{3}+C=-\frac{1}{tgx}=2tgx+\frac{1}{3}\cdot tg^3x+C $$
