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

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(A) ધારો કે $I = \int_0^{\pi /4} {\frac{{\sin x + \cos x}}{{9 + 16\sin 2x}}\,dx}$.$t = \sin x - \cos x$ આદેશ લેતા,$dt = (\cos x + \sin x)dx$ મળે.જ્યાર

Vedclass · Accepted Answer

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

Vedclass · Answer

(A) ધારો કે $I = \int_0^{\pi /4} {\frac{{\sin x + \cos x}}{{9 + 16\sin 2x}}\,dx}$.$t = \sin x - \cos x$ આદેશ લેતા,$dt = (\cos x + \sin x)dx$ મળે.જ્યારે $x = 0$,ત્યારે $t = \sin 0 - \cos 0 = -1$.જ્યારે $x = \pi /4$,ત્યારે $t = \sin(\pi /4) - \cos(\pi /4) = 0$.વળી,$t^2 = (\sin x - \cos x)^2 = 1 - \sin 2x$,તેથી $\sin 2x = 1 - t^2$.આ કિંમતો સંકલનમાં મૂકતા:$I = \int_{-1}^0 \frac{dt}{9 + 16(1 - t^2)} = \int_{-1}^0 \frac{dt}{25 - 16t^2}$.સૂત્ર $\int \frac{dx}{a^2 - x^2} = \frac{1}{2a} \log \left| \frac{a+x}{a-x} \right|$ નો ઉપયોગ કરતા:$I = \frac{1}{16} \int_{-1}^0 \frac{dt}{(5/4)^2 - t^2} = \frac{1}{16} \cdot \frac{1}{2(5/4)} \left[ \log \left| \frac{5/4 + t}{5/4 - t} \right| \right]_{-1}^0$.$I = \frac{1}{40} \left[ \log \left| \frac{5 + 4t}{5 - 4t} \right| \right]_{-1}^0 = \frac{1}{40} \left[ \log(1) - \log \left| \frac{1}{9} \right| \right]$.$I = \frac{1}{40} [0 - (-\log 9)] = \frac{1}{40} \log(3^2) = \frac{2}{40} \log 3 = \frac{1}{20} \log 3$.

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

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