Anal Practice4

Тука сложи заглавие


страницата се нуждае от дописване/преглеждане


Интеграли от вида $I(m,n) = \int \sin ^m x \cos^n x dx, m,n \int \mathfrac{Z}$

Задача 1

(1)
\begin{eqnarray} \int \sin^m x dx = I(m,0) & = & \\ & = & -\int \sin^{m-1} x d cos x = \\ & = & -\cos x \sin^{m-1} x + \int \cos x (\sin^{m-1} x)' dx = \\ & = & -\cos x \sin^{m-1} x + (m-1)\int \cos x \sin^{m-2} x \cos x dx = \\ & = & -\cos x \sin^{m-1} x + (m-1)\int \cos^2 x \sin^{m-2} x dx = \\ & = & -\cos x \sin^{m-1} x + (m-1)\int (1-\sin^2 x) \sin^{m-2} x dx = \\ & = & -\cos x \sin^{m-1} x + (m-1)\int \sin^{m-2} x dx - (m-1) \int \sin^m x dx = \\ & = & -\cos x \sin^{m-1} x + (m-1)\int \sin^{m-2} x dx - m I(m,0) + I(m,0) \end{eqnarray}

В крайна сметка получаваме, че

(2)
\begin{eqnarray} I(m,0)= -\dfrac{1}{m}\cos x \sin^{m-1} x + \dfrac{m-1}{m}I(m-2,0) \end{eqnarray}

Като формулата е вярна за $m > 2$
$\int \sin x dx = -\cos x$
$\int \sin^2 x dx = \int \dfrac{1-\cos 2x}{2}dx = \dfrac{x}{2}-\dfrac{1}{4}\sin 2x$

Задача 2

Просто прилагаме горните формулки

(3)
\begin{eqnarray} \int sin^5 x dx = -\int \sin^4 x d \cos x = -\cos \sin^4 x + 4\int \cos^2 x \sin^3 x dx = \\ = - \cos x \sin^4 x + 4\int \sin^3 x dx - 4 \int \sin^5 x dx \\ I(5,0) = -\dfrac{1}{5} \cos x \sin^4 x + \dfrac{4}{5}I(3,0) = -\dfrac{1}{5} \cos x \sin^4 x - \dfrac{4}{5}\cos x \sin^2 x + \dfrac{8}{15}\cos^3 x \end{eqnarray}

Задача 3

(4)
\begin{align} \int sin^6 x dx \end{align}

Задача 4

(5)
\begin{align} \int sin^4 x dx \end{align}

Задача 5

(6)
\begin{eqnarray} \int \dfrac{1}{ sin^m x} dx = J(m,0) & = & \\ & = & \int \dfrac{\sin^2 x + \cos^2 x}{ sin^m x} dx = \\ & = & \int \dfrac{1}{ sin^{m-2} x} dx + \int \cos^2 x sin^{-m} x dx = \\ & = & J(m-2,0) + \int \cos^2 x sin^{-m} x dx = \\ & = & J(m-2,0) + \dfrac{1}{1-m}\int \cos x d \sin^{1-m} x = \\ & = & J(m-2,0) + \dfrac{1}{1-m}(\cos x \sin^{1-m} x +\int \sin^{1-m} x \sin x d x)= \\ & = & J(m-2,0) + \dfrac{1}{1-m}(\cos x \sin^{1-m} x +\int \sin^{2-m} x d x)= \\ & = &(1-\dfrac{1}{m-1}) J(m-2,0) -\dfrac{1}{m-1}\dfrac{\cos x}{\sin^{m-1} x} \end{eqnarray}

пак искаме $m > 2$ и знаем, че
$\int \dfrac{dx}{\sin^2 x} = -cotg x$ и $\int \dfrac{dx}{\sin x} = \ln |tg \frac{x}{2}|$

Задача 6

(7)
\begin{align} \int \dfrac{dx}{\sin^3 x} = \dfrac{1}{2} \ln |\tg \frac{x}{2}|-\dfrac{1}{2}\dfrac{\cos x}{\sin^2 x} \end{align}

Задача 7

(8)
\begin{align} \int \cos^4 x dx = \int \cos^3 x d \sin x & = & \\ & = & \cos^3 x \sin x - \int \sin x (\cos^3 x)' dx = \\ & = & \cos^3 x \sin x + 3 \int \sin^2 x \cos^2 x dx = \\ \end{align}
Unless otherwise stated, the content of this page is licensed under Creative Commons Attribution-NonCommercial-ShareAlike 3.0 License