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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 formI = \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=0I 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 Ibecomes \dfrac{1}{{\sqrt {{a^2} + {b^2}} }}\int {{\rm{cosec}}\left( {x + \phi } \right)dx}
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