Evaluate _ 0 ^ x a ^ 2 ^ 2 x + b ^ 2 ^ 2 x d x.

Question

Evaluate $ \int _ { 0 } ^ { \pi } \frac { x } { a ^ { 2 } \cos ^ { 2 } x + b ^ { 2 } \sin ^ { 2 } x } d x.$

✓

Answer

According to the question , $ I = \int _ { 0 } ^ { \pi } \frac { x } { a ^ { 2 } \cos ^ { 2 } x + b ^ { 2 } \sin ^ { 2 } x } d x$ ...(i)
$ \Rightarrow I = \int _ { 0 } ^ { \pi } \frac { ( \pi - x ) } { a ^ { 2 } \cos ^ { 2 } ( \pi - x ) + b ^ { 2 } \sin ^ { 2 } ( \pi - x ) } d x$$ \left[ \because \int _ { 0 } ^ { a } f ( x ) d x = \int _ { 0 } ^ { a } f ( a - x ) d x \right]$
$ \Rightarrow \quad I = \int _ { 0 } ^ { \pi } \frac { ( \pi - x ) } { a ^ { 2 } \cos ^ { 2 } x + b ^ { 2 } \sin ^ { 2 } x } d x$ ...(ii)
On adding Equations (i) and (ii) we get ,
$ 2 I = \int _ { 0 } ^ { \pi } \frac { ( x + \pi - x ) } { a ^ { 2 } \cos ^ { 2 } x + b ^ { 2 } \sin ^ { 2 } x } d x$
$ \Rightarrow \quad 2 I = \pi \int _ { 0 } ^ { \pi } \frac { d x } { a ^ { 2 } \cos ^ { 2 } x + b ^ { 2 } \sin ^ { 2 } x }$
we know that,$ \int _ { 0 } ^ { 2 a } f ( x ) d x = 2 \int _ { 0 } ^ { a }f(x) dx \ , \ if \ \ f(2a - x) = f(x)$
Here, $ a^2 cos^2( \pi - x ) + b ^ { 2 } \sin ^ { 2 } ( \pi - x )$
$= a^2 cos^2 x + b^2 sin^2x$
$ \therefore \quad 2 I = 2 \pi \int _ { 0 } ^ { \pi / 2 } \frac { d x } { a ^ { 2 } \cos ^ { 2 } x + b ^ { 2 } \sin ^ { 2 } x }$
On dividing numerator and denominator by $cos^2x$ we get ,
$ 2 I = 2 \pi \int _ { 0 } ^ { \pi / 2 } \frac { \sec ^ { 2 } x } { a ^ { 2 } + b ^ { 2 } \tan ^ { 2 } x } d x$
Put $tan x = t \implies sec^2 x dx = dt$
Lower limit when x = 0, then $t = tan 0 = 0$
Upper limit when x = $ \frac { \pi } { 2 },$then $ t =\tan \frac { \pi } { 2 } = \infty$
$ \therefore \quad I = \pi \int _ { 0 } ^ { \infty } \frac { d t } { a ^ { 2 } + b ^ { 2 } t ^ { 2 } }$
$ = \pi \int _ { 0 } ^ { \infty } \frac { d t } { a ^ { 2 } + ( b t ) ^ { 2 } }$

$ = \frac { \pi } { b ^ { 2 } } \int _ { 0 } ^ { \infty } \frac { d t } { \left( \frac { a } { b } \right) ^ { 2 } + t ^ { 2 } }$
$\Rightarrow I = \frac { \pi } { a b } \left[ \tan ^ { - 1 } \frac { b t } { a } \right] _ { 0 } ^ { \infty } \left[ \because \int \frac { d x } { a ^ { 2 } + x ^ { 2 } } = \frac { 1 } { a } \tan ^ { - 1 } \frac { x } { a } + C \right]$
$\Rightarrow I = \frac { \pi } { a b } \left[ \tan ^ { - 1 } \infty - \tan ^ { - 1 } 0 \right]$
$\Rightarrow I = \frac { \pi } { a b } \left[ \frac { \pi } { 2 } - 0 \right]$$ \left[ \begin{array} { c } { \because \tan ^ { - 1 } \infty = \tan ^ { - 1 } \left( \tan \frac { \pi } { 2 } \right) = \frac { \pi } { 2 } } \\ { \text { and } \tan ^ { - 1 } 0 = \tan ^ { - 1 } \left( \tan 0 ^ { \circ } \right) = 0 } \end{array} \right]$

$ \therefore I = \frac { \pi ^ { 2 } } { 2 a b }$

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