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Integrals — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsIntegrals1 Mark
Question
Evaluate the definite integral $\int_0^{\frac{\pi}{4}} \frac{\sin x+\cos x}{9+16 \sin 2 x} d x$
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Answer

According to the question, $I = \int _ { 0 } ^ { \pi / 4 } \frac { \sin x + \cos x } { 9 + 16 \sin 2 x } d x$
$\Rightarrow \quad I = \int _ { 0 } ^ { \pi / 4 } \frac { \sin x + \cos x } { 9 + 16 ( 1 + \sin 2 x - 1 ) } d x$
$\Rightarrow \quad I = \int _ { 0 } ^ { \pi / 4 } \frac { \sin x + \cos x } { 9 + 16 [ 1 - ( 1 - \sin 2 x ) ] } d x$
$\Rightarrow I = \int _ { 0 } ^ { \pi / 4 } \frac { \sin x + \cos x } { 9 + 16 \left[ \begin{array} { r } { 1 - \left( \cos ^ { 2 } x + \sin ^ { 2 } x \right. } { - 2 \sin x \cos x ) } \end{array} \right] } d x$$[\because 1 = \cos ^ { 2 } x + \sin ^ { 2 } x ]$ and $[\because \sin 2 x = 2 \sin x \cos x ]$
$\Rightarrow \quad I = \int _ { 0 } ^ { \pi / 4 } \frac { \sin x + \cos x } { 9 + 16 \left[ 1 - ( \cos x - \sin x ) ^ { 2 } \right] } d x$
put , $cos x - sin x = t$
$\Rightarrow (- sin x - cos x) dx = dt$ 
$\Rightarrow (sin x + cos x) dx = - dt$ 
Lower limit , when x = 0, then t = cos 0 - sin 0 = 1 
Upper limit , when x = $\frac { \pi } { 4 } , $ then $t = \cos \frac { \pi } { 4 } - \sin \frac { \pi } { 4 } = \frac { 1 } { \sqrt { 2 } } - \frac { 1 } { \sqrt { 2 } } = 0.$
$\therefore \quad I = \int _ { 1 } ^ { 0 } \frac { - d t } { 9 + 16 \left( 1 - t ^ { 2 } \right) }$
$\Rightarrow \quad I = \int _ { 0 } ^ { 1 } \frac { d t } { 9 + 16 \left( 1 - t ^ { 2 } \right) }$
$= \int _ { 0 } ^ { 1 } \frac { d t } { 25 - 16 t ^ { 2 } }$
$= \frac { 1 } { 16 } \int _ { 0 } ^ { 1 } \frac { d t } { \left( \frac { 5 } { 4 } \right) ^ { 2 } - t ^ { 2 } }$
$= \frac { 1 } { 2 \times \frac { 5 } { 4 } \times 16 } \left[ \log \left| \frac { 5 + 4 t } { 5 - 4 t } \right| \right] _ { 0 } ^ { 1 }$$\left[ \because \int \frac { 1 } { a ^ { 2 } - x ^ { 2 } } d x = \frac { 1 } { 2 a } \log \left| \frac { a + x } { a - x } \right| + C \right]$
$= \frac { 1 } { 40 } \left[ \log \left| \frac { 5 + 4 } { 5 - 4 } \right| - \log \left| \frac { 5 } { 5 } \right| \right]$
$= \frac { 1 } { 40 } \left[ \log \left( \frac { 9 } { 1 } \right) - \log \left( \frac { 5 } { 5 } \right) \right]$
$= \frac { 1 } { 40 } ( \log 9 - \log 1 ) $
$= \frac { 1 } { 40 } ( \log 9 )$ [$ \because$ log 1 = 0]
$\Rightarrow I = \frac { 1 } { 40 } \log ( 3 ) ^ { 2 }$
$= \frac { 2 } { 40 } \log 3$ [$\because$log an = nlog a]
$\therefore \quad I = \frac { 1 } { 20 } \log 3$

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