### Definite integration(10.4-10.6)

#### Use integration by substitution to find definite integrals

• Evaluate $$\int_{\; - \,\frac{\pi }{3}}^{\;\frac{\pi }{3}} {\;\frac{{dx}}{{{{\cos }^2}x\sqrt {4 - {{\tan }^2}x} }}}$$.
• ∵ $$\frac{1}{{{{\cos }^2}( - x)\sqrt {4 - {{\tan }^2}( - x)} }} = \frac{1}{{{{\cos }^2}x\sqrt {4 - {{\tan }^2}x} }}$$
∴ $$\frac{1}{{{{\cos }^2}x\sqrt {4 - {{\tan }^2}x} }}$$ is an even function.
$$\int_{\; - \,\frac{\pi }{3}}^{\;\frac{\pi }{3}} {\;\frac{{dx}}{{{{\cos }^2}x\sqrt {4 - {{\tan }^2}x} }}} = 2\int_{\;0}^{\;\frac{\pi }{3}} {\,\frac{{{{\sec }^2}x\,dx}}{{\sqrt {4 - {{\tan }^2}x} }}}$$
Let $$u = \tan x$$, then $$du = {\sec ^2}x\,dx$$.
When $$x = 0$$, $$u = 0$$.
When $$x = \frac{\pi }{3}$$, $$u = \sqrt 3$$.
$$\int_{\; - \,\frac{\pi }{3}}^{\;\frac{\pi }{3}} {\;\frac{{dx}}{{{{\cos }^2}x\sqrt {4 - {{\tan }^2}x} }}} = 2\int_{\;0}^{\;\sqrt 3 } {\,\frac{{du}}{{\sqrt {4 - {u^2}} }}}$$
Let $$u = 2\sin \theta$$, where $$- {\rm{ }}\frac{\pi }{2} < \theta < \frac{\pi }{{\rm{2}}}$$,
then $$du = {\rm{2}}\cos \theta \,d{\kern 1pt} \theta$$ and $$\sqrt {4 - {u^2}} = 2\cos \theta$$.
When $$u = 0$$, $$\theta = 0$$.
When $$u = \sqrt 3$$, $$\theta = \frac{\pi }{3}$$.
$$\begin{array}{1}\int_{\; - \,\frac{\pi }{3}}^{\;\frac{\pi }{3}} {\;\frac{{dx}}{{{{\cos }^2}x\sqrt {4 - {{\tan }^2}x} }}} = 2\int_{\;0}^{\;\frac{\pi }{3}} {\,\frac{{2\cos \theta \,d{\kern 1pt} \theta }}{{2\cos \theta }}} \\ = 2\int_{\;0}^{\;\frac{\pi }{3}} {\,d\theta } \\ = 2\,[\theta ]{\kern 1pt} _0^{\frac{\pi }{3}}\\ = {\underline{\underline {\frac{{2\pi }}{3}}} ^{^{}}}\end{array}$$

#### Use integration by parts to find definite integrals

• Evaluate $$\int_{\;0}^{\;3} {\,{{(\frac{x}{{{e^{2x}}}})}^2}dx}$$.
• $$\begin{array}{1}\int_{\;0}^{\;3} {\,{{(\frac{x}{{{e^{2x}}}})}^2}dx} = \int_{\;0}^{\;3} {\,{x^2}{e^{ - 4x}}\,dx} \\ = - {\rm{ }}\frac{1}{4}\int_{\;0}^{\;3} {\,{x^2}\,d({e^{ - 4x}})} \\ = - {\rm{ }}\frac{1}{4}[{x^2}{e^{ - 4x}}]{\kern 1pt} _0^3\, + \frac{1}{4}\int_{\;0}^{\;3} {\,{e^{ - 4x}}\,d({x^2})} \\ = - {\rm{ }}\frac{1}{4}(9{e^{ - 12}} - 0) + \frac{1}{2}\int_{\;0}^{\;3} {\,x{e^{ - 4x}}\,dx} \\ = - {\rm{ }}\frac{9}{4}{e^{ - 12}} - \frac{1}{8}\int_{\;0}^{\;3} {\,xd({e^{ - 4x}})} \\ = - {\rm{ }}\frac{9}{4}{e^{ - 12}} - \frac{1}{8}[x{e^{ - 4x}}]{\kern 1pt} _0^3\, + \frac{1}{8}\int_{\;0}^{\;3} {\,{e^{ - 4x}}\,dx} \\ = - {\rm{ }}\frac{9}{4}{e^{ - 12}} - \frac{1}{8}(3{e^{ - 12}} - 0) - \frac{1}{{32}}[{e^{ - 4x}}]{\kern 1pt} _0^3\\ = \underline{\underline { - {\rm{ }}\frac{{85}}{{32}}{e^{ - 12}} + \frac{1}{{32}}}} \end{array}$$

#### Understand the properties of the definite integrals of even, odd and periodic functions

• [TheoryMissing]
• Let $$f(x)$$, $$g(x)$$ and $$h(x)$$ be odd functions, $$p(x)$$ and $$q(x)$$ be even functions.
(a) Prove that the following functions are odd functions.
(i) $$h(g(f(x)))$$ (ii) $$f(g(x)) \cdot p(q(x))$$
(b) Evaluate the following definite integrals.
(i) $$\int_{\; - \,\frac{\pi }{2}}^{\;\frac{\pi }{2}} {\;{{\sin }^3}(2{x^5})\,dx}$$ (ii) $$\int_{\; - \pi }^{\;\pi } {\,\cos {x^2} \cdot \sqrt[3]{{x - \sin x}}\,dx}$$
• (a) (i)
∵ $$\begin{array}{1}h{\kern 1pt} (g(f( - x))) = h{\kern 1pt} (g( - f(x)))\\ = h{\kern 1pt} ( - {\rm{ }}g(f(x)))\\ = - {\rm{ }}h{\kern 1pt} (g(f(x)))\end{array}$$
∴ $$h{\kern 1pt} (g(f(x)))$$ is an odd function.
(ii)
∵ $$\begin{array}{1}f(g( - x)) \cdot p(q( - x)) = f( - g(x)) \cdot p(q(x))\\ = - f(g(x)) \cdot p(q(x))\end{array}$$
∴ $$f(g(x)) \cdot p(q(x))$$ is an odd function.

(b) (i)
∵ $$\begin{array}{1}2{( - x)^5} = - {\rm{ }}2{x^5}\ \\sin ( - x) = - {\rm{ }}\sin x \\{( - x)^3} = - {x^3}\end{array}$$
∴ Take $$f(x) = 2{x^5}$$, $$g(x) = \sin x$$ and $$h(x) = {x^3}$$,
$$\begin{array}{1}h(g(f(x))) = {[\sin (2{x^5})]^3}\\ = {\sin ^3}(2{x^5})\end{array}$$
∴ $${\sin ^3}(2{x^5})$$ is an odd function.
Hence $$\int_{\; - \,\frac{\pi }{2}}^{\;\frac{\pi }{2}} {\;{{\sin }^3}(2{x^5})\,dx} = \underline{\underline 0}$$.
(ii)
∵ $$\begin{array}{1}\sqrt[3]{{ - {\rm{ }}x}} = - {\rm{ }}\sqrt[3]{x}\\ - {\rm{ }}x - \sin ( - x) = - {\rm{ }}{(x - \sin x)^{}}\\ \cos ( - x) = \cos {x^{}}\\ {( - x)^2} = {x^2}\end{array}$$
∴ Take $$f(x) = \sqrt[3]{x}$$, $$g(x) = x - \sin x$$, $$p(x) = \cos x$$ and $$q(x) = {x^2}$$,
$$\begin{array}{1}f(g(x)) \cdot p(q(x)) = \sqrt[3]{{x - \sin x}} \cdot \cos ({x^2})\\ = \cos {x^2} \cdot {\sqrt[3]{{x - \sin x}}^{}}\end{array}$$
∴ $$\cos {x^2} \cdot \sqrt[3]{{x - \sin x}}$$ is an odd function.
Hence $$\int_{\; - \pi }^{\;\pi } {\,\cos {x^2} \cdot \sqrt[3]{{x - \sin x}}dx} = \underline{\underline 0}$$.