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# 数学代写|微积分代写Calculus代考|MTH251

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## 数学代写|微积分代写Calculus代考|Integration by Parts

Integration by parts is a technique for simplifying integrals of the form
$$\int u(x) v^{\prime}(x) d x .$$
It is useful when $u$ can be differentiated repeatedly and $v^{\prime}$ can be integrated repeatedly without difficulty. The integrals
$$\int x \cos x d x \text { and } \int x^2 e^x d x$$
are such integrals because $u(x)=x$ or $u(x)=x^2$ can be differentiated repeatedly to become zero, and $v^{\prime}(x)=\cos x$ or $v^{\prime}(x)=e^x$ can be integrated repeatedly without difficulty. Integration by parts also applies to integrals like
$$\int \ln x d x \text { and } \int e^x \cos x d x .$$
In the first case, the integrand $\ln x$ can be rewritten as $(\ln x)(1)$, and $u(x)=\ln x$ is easy to differentiate while $v^{\prime}(x)=1$ easily integrates to $x$. In the second case, each part of the integrand appears again after repeated differentiation or integration.
Product Rule in Integral Form
If $u$ and $v$ are differentiable functions of $x$, the Product Rule says that
$$\frac{d}{d x}[u(x) v(x)]=u^{\prime}(x) v(x)+u(x) v^{\prime}(x) .$$

In terms of indefinite integrals, this equation becomes
$$\int \frac{d}{d x}[u(x) v(x)] d x=\int\left[u^{\prime}(x) v(x)+u(x) v^{\prime}(x)\right] d x$$
or
$$\int \frac{d}{d x}[u(x) v(x)] d x=\int u^{\prime}(x) v(x) d x+\int u(x) v^{\prime}(x) d x .$$
Rearranging the terms of this last equation, we get
$$\int u(x) v^{\prime}(x) d x=\int \frac{d}{d x}[u(x) v(x)] d x-\int v(x) u^{\prime}(x) d x,$$
leading to the integration by parts formula
Integration by Parts Formula
$$\int u(x) v^{\prime}(x) d x=u(x) v(x)-\int v(x) u^{\prime}(x) d x$$

## 数学代写|微积分代写Calculus代考|Trigonometric Integrals

Trigonometric integrals involve algebraic combinations of the six basic trigonometric functions. In principle, we can always express such integrals in terms of sines and cosines, but it is often simpler to work with other functions, as in the integral
$$\int \sec ^2 x d x=\tan x+C .$$
The general idea is to use identities to transform the integrals we have to find into integrals that are easier to work with.
Products of Powers of Sines and Cosines
We begin with integrals of the form
$$\int \sin ^m x \cos ^n x d x,$$
where $m$ and $n$ are nonnegative integers (positive or zero). We can divide the appropriate substitution into three cases according to $m$ and $n$ being odd or even.

Case 1 If $\boldsymbol{m}$ is odd, we write $m$ as $2 k+1$ and use the identity $\sin ^2 x=$ $1-\cos ^2 x$ to obtain
$$\sin ^m x=\sin ^{2 k+1} x=\left(\sin ^2 x\right)^k \sin x=\left(1-\cos ^2 x\right)^k \sin x .$$
Then we combine the single $\sin x$ with $d x$ in the integral and set $\sin x d x$ equal to $-d(\cos x)$.

Case 2 If $\boldsymbol{n}$ is odd in $\int \sin ^m x \cos ^n x d x$, we write $n$ as $2 k+1$ and use the identity $\cos ^2 x=1-\sin ^2 x$ to obtain
$$\cos ^n x=\cos ^{2 k+1} x=\left(\cos ^2 x\right)^k \cos x=\left(1-\sin ^2 x\right)^k \cos x .$$
We then combine the single $\cos x$ with $d x$ and set $\cos x d x$ equal to $d(\sin x)$.
Case 3 If both $\boldsymbol{m}$ and $\boldsymbol{n}$ are even in $\int \sin ^m x \cos ^n x d x$, we substitute
$$\sin ^2 x=\frac{1-\cos 2 x}{2}, \quad \cos ^2 x=\frac{1+\cos 2 x}{2}$$
to reduce the integrand to one in lower powers of $\cos 2 x$.

## 数学代写|微积分代写Calculus代考|Integration by Parts

$$\int u(x) v^{\prime}(x) d x .$$

$$\int x \cos x d x \text { and } \int x^2 e^x d x$$

$$\int \ln x d x \text { and } \int e^x \cos x d x .$$

$$\frac{d}{d x}[u(x) v(x)]=u^{\prime}(x) v(x)+u(x) v^{\prime}(x) .$$

$$\int \frac{d}{d x}[u(x) v(x)] d x=\int\left[u^{\prime}(x) v(x)+u(x) v^{\prime}(x)\right] d x$$

$$\int \frac{d}{d x}[u(x) v(x)] d x=\int u^{\prime}(x) v(x) d x+\int u(x) v^{\prime}(x) d x .$$

$$\int u(x) v^{\prime}(x) d x=\int \frac{d}{d x}[u(x) v(x)] d x-\int v(x) u^{\prime}(x) d x,$$

$$\int u(x) v^{\prime}(x) d x=u(x) v(x)-\int v(x) u^{\prime}(x) d x$$

## 数学代写|微积分代写Calculus代考|Trigonometric Integrals

$$\int \sec ^2 x d x=\tan x+C .$$

sin和cos的幂的乘积

$$\int \sin ^m x \cos ^n x d x,$$

## MATLAB代写

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