જો int log left(6 sin ^2 x+17 sin x+12right)^ | vedclass

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Question

(D) ધારો કે $I = \int \log (6 \sin^2 x + 17 \sin x + 12)^{\cos x} dx$.$\sin x = t$ લેતા,$\cos x dx = dt$ મળે.તેથી $I = \int \log (6t^2 + 17t + 12) dt

Vedclass · Accepted Answer

$\frac{1}{6}[15 \log 5+14 \log 7-29]$

Vedclass · Answer

(D) ધારો કે $I = \int \log (6 \sin^2 x + 17 \sin x + 12)^{\cos x} dx$.$\sin x = t$ લેતા,$\cos x dx = dt$ મળે.તેથી $I = \int \log (6t^2 + 17t + 12) dt = \int \log ((2t+3)(3t+4)) dt = \int (\log(2t+3) + \log(3t+4)) dt$.ખંડશઃ સંકલન (Integration by parts) $\int u dv = uv - \int v du$ નો ઉપયોગ કરતા:$\int \log(2t+3) dt = t \log(2t+3) - \int \frac{2t}{2t+3} dt = t \log(2t+3) - \int (1 - \frac{3}{2t+3}) dt = (t + \frac{3}{2}) \log(2t+3) - t$.તે જ રીતે,$\int \log(3t+4) dt = (t + \frac{4}{3}) \log(3t+4) - t$.આમ,$f(t) = (t + \frac{3}{2}) \log(2t+3) + (t + \frac{4}{3}) \log(3t+4) - 2t$.$x = \frac{\pi}{2}$ માટે,$t = \sin(\frac{\pi}{2}) = 1$.$f(1) = (1 + \frac{3}{2}) \log(5) + (1 + \frac{4}{3}) \log(7) - 2(1) = \frac{5}{2} \log 5 + \frac{7}{3} \log 7 - 2 = \frac{15 \log 5 + 14 \log 7 - 12}{6}$.

જો $\int \log \left(6 \sin ^2 x+17 \sin x+12\right)^{\cos x} d x=f(x)+c$ હોય,તો $f\left(\frac{\pi}{2}\right)=$

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