ONE MORE STEP TOWARD BETTER TOMORROW : CHAPTER 9

Monday 4 August 2014

CHAPTER 9 - MORE Worked Out Examples

Example: 1

Evaluate the following integrals:

	(a) $\int {\dfrac{1}{{1 + \sin x}}} \,\,dx$
	(b) $\int {\dfrac{1}{{1 + \sin x + \cos x}}} \,\,dx$

Solution: 1-(a)

In both these examples, the substitutions that we have seen uptill now will not work. (You are urged to try it for yourself).

For such functions, we use a new type of substitution. We use the following half – angle formulae:

	$\sin x = \dfrac{{2\tan x/2}}{{1 + {{\tan }^2}x/2}}$
	$\cos x = \dfrac{{1 - {{\tan }^2}x/2}}{{1 + {{\tan }^2}x/2}}$

Making these substitutions converts the given integrals into a form where the substitution $t = \tan \dfrac{x}{2}$ is possible. Lets apply it on the current example.

	$I = \int {\dfrac{1}{{1 + \sin x}}} \,\,dx$
	$= \int {\dfrac{1}{{1 + \dfrac{{2\tan x/2}}{{1 + {{\tan }^2}x/2}}}}} \,\,dx$
	$= \int {\dfrac{{1 + {{\tan }^2}x/2}}{{1 + 2\tan x/2 + {{\tan }^2}x/2}}} \,\,dx$
	$= \int {\dfrac{{{{\sec }^2}x/2}}{{{{\left( {1 + \tan x/2} \right)}^2}}}} \,\,dx$

We now substitute $\tan \dfrac{x}{2} = t$ . Thus, $\dfrac{1}{2}{\sec ^2}\dfrac{x}{2}dx = dt$

	$\Rightarrow \,\,\,\, I = 2\int {\dfrac{{dt}}{{{{\left( {1 + t} \right)}^2}}}}$
	$= \dfrac{{ - 2}}{{1 + t}} + C$
	$= \dfrac{{ - 2}}{{1 + \tan x/2}} + C$

Solution: 1-(b)

	$I = \int {\dfrac{1}{{1 + \sin x + \cos x}}} \,\,dx$
	$= \int {\dfrac{1}{{1 + \dfrac{{2\tan x/2}}{{1 + {{\tan }^2}x/2}} + \dfrac{{1 - {{\tan }^2}x/2}}{{1 + {{\tan }^2}x/2}}}}} \,\,dx$
	$= \int {\dfrac{{{{\sec }^2}x/2}}{{2 + 2\tan x/2}}} \,\,dx$

Substitute $\tan \dfrac{x}{2} = t$ . Thus, ${\sec ^2}\dfrac{x}{2}dx = 2dt$

	$\Rightarrow \,\,\,\, I = \int {\dfrac{{dt}}{{1 + t}}}$
	$= \ln \left\| {1 + t} \right\| + C$
	$= \ln \left\| {1 + \tan \dfrac{x}{2}} \right\| + C$

These two examples should make it clear that a general integral of the form $I = \int {\dfrac{1}{{a\sin x + b\cos x + c}}} \,\,dx$ can always be integrated using the mentioned substitutions. Let us analyse the general case itself.

	$I = \int {\dfrac{1}{{a\sin x + b\cos x + c}}} \,\,dx$	$\ldots(1)$
	$= \int {\dfrac{1}{{\dfrac{{2a\tan x/2}}{{1 + {{\tan }^2}x/2}} + \dfrac{{b - b{{\tan }^2}x/2}}{{1 + {{\tan }^2}x/2}} + c}}} \,\,dx$
	$= \int {\dfrac{{{{\sec }^2}x/2}}{{\left( {c - b} \right){{\tan }^2}x/2 + 2a\tan x/2 + \left( {c + b} \right)}}} \,\,dx$

The substitution $\tan \dfrac{x}{2} = t$ reduces this integral to

	$I = 2\int {\dfrac{{dt}}{{\left( {c - b} \right){t^2} + 2at + \left( {c + b} \right)}}}$
	$= 2\int {\dfrac{{dt}}{{A{t^2} + Bt + C}}}$

We have already evaluated integrals of this form. They correspond to either $(17)$ or $(21)$ .

In ( $1$ ) above, if $c=0$ , $I$ can also be solved by writing $\left( {a\sin x + b\cos x} \right) = \sqrt {{a^2} + {b^2}} \,\sin \left( {x + \phi } \right) \left\{ {\phi = {{\tan }^{ - 1}}\dfrac{b}{a}} \right\}$ so that $I$ becomes $\dfrac{1}{{\sqrt {{a^2} + {b^2}} }}\int {{\rm{cosec}}\left( {x + \phi } \right)dx}$

ONE MORE STEP TOWARD BETTER TOMORROW

Monday 4 August 2014

CHAPTER 9 - MORE Worked Out Examples

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