Question

Применяя указанные подстановки, найти следующие интегралы:
1) ∫ dx/3x-4,t=3x-4
2) ∫ dx/√(a^2-x^2),t=ax
3) ∫ 4dx/sin^2*2x,t=2x
4) ∫ sin^3xcosxdx,t=sinx
5) ∫ sin^3xdx,t=cosx
6) ∫ tg xdx, t=cos x
7) ∫ xdx/√(1-x^2),t=1-x^2 или x=sin t
8) ∫ xdx/√(1+x^2), t=1+x^2
9)x^3dx/√(1+x^2), t=1+x^2

Спасибо.

Answer 1

$\displaystyle\int\frac{dx}{3x-4}=\frac{1}{3}\int\frac{dt}{t}=\frac{1}{3}ln|t|+C=\frac{1}{3}ln|3x-4|+C\\\\t=3x-4;dt=3dx;dx=\frac{dt}{3}$

$\displaystyle\int\frac{dx}{a^2-x^2}=\frac{1}{a}\int\frac{dt}{a^2-\frac{t^2}{a^2}}=-a\int\frac{dt}{t^2-a^4}=-\frac{a}{2a^2}ln|\frac{t-a^2}{t+a^2}|+C\\=-\frac{1}{2a}ln|\frac{x-a}{x+a}|+C\\t=ax;x=\frac{t}{a};dx=\frac{dt}{a}$

$\displaystyle\int\frac{4dx}{sin^22x}=2\int\frac{dt}{sin^2t}=-2ctgt+C=-2ctg2x+C\\t=2x;dt=2dx;dx=\frac{dt}{2}$

$\displaystyle\int sin^3xcosxdx=\int t^3dt=\frac{t^4}{4}+C=\frac{sin^4x}{4}+C\\t=sinx;dt=cosxdx;dx=\frac{dt}{cosx}$

$\displaystyle\int sin^3xdx=\int(t^2-1)dt=\frac{t^3}{3}-t+C=\frac{cos^3x}{3}-cosx+C\\t=cosx;dt=-sinxdx;dx=-\frac{dt}{sinx}$

$\displaystyle\int tg xdx=-\int\frac{dt}{t}=-ln|t|+C=-ln|cosx|+C\\t=cosx;dt=-sinxdx;dx=-\frac{dt}{sinx}$

$\displaystyle\int\frac{xdx}{\sqrt{1-x^2}}=-\frac{1}{2}\int\frac{dt}{\sqrt t}=-\sqrt t+C=-\sqrt{1-x^2}+C\\t=1-x^2;dt=-2xdx;dx=-\frac{dt}{2x}\\\\\int\frac{xdx}{\sqrt{1-x^2}}=\int sintdt=-cost+C=-cos(arcsinx)+C\\x=sint;t=arcsinx;dt=\frac{dx}{\sqrt{1-x^2}};dx=\sqrt{1-x^2}dt$

$\displaystyle \int\frac{xdx}{\sqrt{1+x^2}}=\frac{1}{2}\int\frac{dt}{\sqrt t}=\sqrt t+C=\sqrt{1+x^2}+C\\t=1+x^2;dt=2xdx;dx=\frac{dt}{2x}$

$\displaystyle \int\frac{x^3dx}{\sqrt{1+x^2}}=\frac{1}{2}\int\frac{t-1}{\sqrt t}dt=\frac{1}{2}\int(\sqrt t-\frac{1}{\sqrt t})dt=\frac{\sqrt{t^3}}{3}-\sqrt t+C\\=\frac{\sqrt{(1+x^2)^3}}{3}-\sqrt{1+x^2}+C\\t=1+x^2;x^2=t-1;dt=2xdx;dx=\frac{dt}{2x}$