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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 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 $\dfrac{1}{{\sqrt {{a^2} + {b^2}} }}\int {{\rm{cosec}}\left( {x + \phi } \right)dx}$