int_0^{pi /4} {frac{{sin x + cos x}}{{9 + 16s | vedclass

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Question

(A) Let $I = \int_0^{\pi /4} {\frac{{\sin x + \cos x}}{{9 + 16\sin 2x}}\,dx}$.Substitute $t = \sin x - \cos x$,then $dt = (\cos x + \sin x)dx$.When $x

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$\frac{1}{{20}}\log 3$

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(A) Let $I = \int_0^{\pi /4} {\frac{{\sin x + \cos x}}{{9 + 16\sin 2x}}\,dx}$.Substitute $t = \sin x - \cos x$,then $dt = (\cos x + \sin x)dx$.When $x = 0$,$t = \sin 0 - \cos 0 = -1$.When $x = \pi /4$,$t = \sin(\pi /4) - \cos(\pi /4) = 0$.Also,$t^2 = (\sin x - \cos x)^2 = \sin^2 x + \cos^2 x - 2\sin x \cos x = 1 - \sin 2x$,so $\sin 2x = 1 - t^2$.Substituting these into the integral:$I = \int_{-1}^0 \frac{dt}{9 + 16(1 - t^2)} = \int_{-1}^0 \frac{dt}{9 + 16 - 16t^2} = \int_{-1}^0 \frac{dt}{25 - 16t^2}$.Using the formula $\int \frac{dx}{a^2 - x^2} = \frac{1}{2a} \log \left| \frac{a+x}{a-x} \right|$,we have:$I = \frac{1}{16} \int_{-1}^0 \frac{dt}{(5/4)^2 - t^2} = \frac{1}{16} \cdot \frac{1}{2(5/4)} \left[ \log \left| \frac{5/4 + t}{5/4 - t} \right| \right]_{-1}^0$.$I = \frac{1}{16} \cdot \frac{2}{5} \left[ \log \left| \frac{5 + 4t}{5 - 4t} \right| \right]_{-1}^0 = \frac{1}{40} \left[ \log(1) - \log \left| \frac{5 - 4}{5 + 4} \right| \right]$.$I = \frac{1}{40} [0 - \log(1/9)] = \frac{1}{40} \log(9) = \frac{1}{40} \log(3^2) = \frac{2}{40} \log 3 = \frac{1}{20} \log 3$.

$\int_0^{\pi /4} {\frac{{\sin x + \cos x}}{{9 + 16\sin 2x}}\,dx = } $

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