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Home  >>  CBSE XII  >>  Math  >>  Model Papers
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Evlauate$ \Large \int_0^{\pi/4}$$ \large \frac{\sin x+\cos x}{9+16\sin 2x}$$dx$

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$ \Large \int_0^{\pi/4}$$ \large \frac{\sin x+\cos x}{9+16\sin 2x}$$dx$
Let $\sin x - \cos x =t$
Differentiating both the sides we get
$(cosx+sinx)dx=dt$
Squaring $t$ we get
$t^2 = \sin^2 x + \cos^2 x - 2\sin x \cos x = 1 - 2 \sin x \cos x = 1 - \sin 2 x $
$\Rightarrow\:sin\;2x=1-t^2$
when $x=0,\:\:t=-1$ and when $x=\large\frac{\pi}{4},\:\:t=0$
$\Rightarrow I = \large \int^0_{-1} \frac{1}{9 + 16 (1-t^2)}$ $dt = \large \int^0_{-1} \frac{1}{25 - 16t^2}$$dt$
$\Rightarrow I = \large\frac{1}{16} \int^0_{-1} \frac {dt}{(5/4)^2-t^2}$
$\Rightarrow I = \large\frac{1}{16}$$ [ -2/5 (\log (5-4t) - \log (4t + 5))]_{-1}^{0} $
$\Rightarrow\: I = \bigg [\large\frac{1}{40} $$[\log (4t + 5) - \log (5-4t)]$$\bigg]_{0}^{-1}$ $ = \large\frac{1}{40}$$ \log 9$
$=\large\frac{1}{20}$$log3$
answered Mar 20, 2014 by balaji.thirumalai
edited Mar 23, 2014 by rvidyagovindarajan_1
 

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