Integration by Parts

Integration by Parts

Theorem 3..5   Let $f(x),g(x)$ be differentiable functions. Then.

$$\int f(x)g^{\prime}(x) = f(x)g(x) - \int f^{\prime}(x)g(x)dx$$

NOTE $(f(x)g(x))' = f'(x)g(x) + f(x)g'(x)$. Then $f(x)g'(x) = (f(x)g(x))' - f'(x)g(x)$. Now integrate both sides with respect $x$.

$$\int f(x)g'(x)dx = f(x)g(x) - \int f'(x)g(x)dx.$$

If $f(x)g'(x)dx$ is hard to integrate, Rewrite by parts to get integration of $\int f(x)g'(x)dx$.

Example 3..4   Integrate the following function by parts.
1. $${\int \log{x} dx}$$ 2. $${\int \sin^{-1}{x}dx}$$

SOLUTION 1. $\left\{\begin{array}{lcl} f(x) = \log{x} & & g'(x) = 1\\ &\searrow& \\ f'(x) = \frac{1}{x} &\leftarrow& g(x) = \int dx = x \end{array}\right.$
Thus

$$\int \underbrace{\log{x}}_{f} \underbrace{1}_{g'}dx = \underbrace{x}_{g} \underbrace{\log{x}}_{f} - \int \underbrace{x}_{g} \underbrace{\frac{1}{x}}_{f'} dx$$

$$= x \log{x} - x + c\ensuremath{\ \blacksquare}$$

2. $\left\{\begin{array}{lcl} f(x) = \sin^{-1}{x} & & g'(x) = dx\\ &\searrow& \\... ...x) = \frac{1}{\sqrt{1-x^2}} &\leftarrow& g(x) = \int dx = x \end{array}\right.$
Thus

$$\int \underbrace{\sin^{-1}{x}}_{f}\underbrace{dx}_{g'} = \underbr... ...}}_{f} - \int \underbrace{x}_{g} \underbrace{\frac{1}{\sqrt{1-x^{2}}}dx}_{f'}.$$

Now let $t = 1 - x^2$. Then $dt = -2x dx$. Thus

$$\int \frac{x}{\sqrt{1-x^{2}}}dx = -\int \frac{dt/2}{\sqrt{t}} = -\frac{1}{2}\int t^{-1/2} dt$$

$$= - \frac{1}{2}2. t^{\frac{1}{2}} = - t^{1/2} + c = - \sqrt{1 - x^2} + c.$$

Thus,

$$\int \sin^{-1}{x} dx = x\sin^{-1}{x} + \sqrt{1 - x^{2}} + c\ensuremath{\ \blacksquare}$$

Exercise 3..4   Integrate the following functions.
1. $${\int x e^{-x} dx}$$ 2. $${\int x\log{x}\:dx}$$ 3. $${\int e^{x}\sin{x}dx}$$

SOLUTION 1. $\left\{\begin{array}{lcl} f(x) = x & & g'(x) = e^{-x}\\ &\searrow& \\ f'(x) = 1 &\leftarrow& g(x) = \int e^{-x}dx = -e^{-x} \end{array}\right.$

Thus,

$$\int xe^{-x} = -xe^{-x} - \int -e^{-x}\: dx$$

$$= -xe^{-x} -e^{-x} + c\ensuremath{\ \blacksquare}$$

2. $\left\{\begin{array}{lcl} f(x) = \log{x} & & g'(x) = x\\ &\searrow& \\ f'(x) = \frac{1}{x} &\leftarrow& g(x) = \int xdx = \frac{x^2}{2} \end{array}\right.$

Thus,

$$\int x\log{x} = \frac{1}{2}x^2 \log{x} - \int \frac{1}{x}\cdot \frac{x^2}{2}\: dx$$

$$= \frac{1}{2}x^2 \log{x} - \int \frac{x}{2}\: dx$$

$$= \frac{1}{2}x^2 \log{x} - \frac{x^2}{4} + c\ensuremath{\ \blacksquare}$$

3.

$$I = \int e^{x}\sin{x}\:dx \hskip 1cm \left(\begin{array}{lcl} f_1 = e... ... \\ f_1' = e^x &\leftarrow& g_1 = \int \sin{x}dx = -\cos{x} \end{array}\right)$$

$$= -e^{x} \cos{x} + \int e^{x} \cos{x}\:dx$$

Now the integral of the second tern is very similar to the original integral $I$. So, we integrate again. .

$$I = -e^{x} \cos{x} + \int e^{x} \cos{x}\:dx \left(\begin{array}{lcl} ... ...s{x}\\ &\searrow& \\ f_2' = e^x &\leftarrow& g_2 = \sin{x} \end{array}\right)$$

$$= -e^{x}\cos{x} + e^{x}\sin{x} - \int e^x \sin{x}\:dx$$

$$= e^{x}(\sin{x} - \cos{x}) - I.$$

Then,

$$2I = e^{x}(\sin{x} - \cos{x}).$$

Thus,

$$I = \frac{1}{2}e^{x}(\sin{x} - \cos{x})\ensuremath{\ \blacksquare}$$

Exercise A

1.

Work out the following integrals?D

(a) $${\int{xe^{x}}\ dx}$$ (b) $${\int{x\sin{x}}\ dx}$$ (c) $${\int{xe^{2x}}\ dx}$$ (d) $${\int{x^{2}e^{x}}\ dx}$$

(e) $${\int{x^{2}\sin{x}}\ dx}$$ (f) $${\int{x^{5}e^{x^{3}}}\ dx}$$ (g) $${\int{x\cos{x}}\ dx}$$

Exercise B

1.

Work out the following integrals?D

(a) $${\int{x\log{x}}\ dx}$$ (b) $${\int{x^2e^{-x}}\ dx}$$ (c) $${\int{(\log{x})^2}\ dx}$$

(d) $${\int{x(x+5)^{14}}\ dx}$$ (e) $${\int{x^{2}\cos{x}}\ dx}$$ (f) $${\int{e^{x}\sin{x}}\ dx}$$

(g) $${\int{\log{(1+x^2)}}\ dx}$$ (h) $${\int{x\tan^{-1}{x}}\ dx}$$ (i) $${\int{x^n\log{x}}\ dx}$$ (n is integer)

(j) $${\int{x^3\sin{x}}\ dx}$$ (k) $${\int{x\sinh{x}}\ dx}$$