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Find the integral of log(sinx+cosx) from -pi/4 to pi/4.

\(\begin{array}{l}I = int_{-pi/4}^{pi/4}log(sinx+cosx)dx\\end{array} \) \(\begin{array}{l}I = int_{-pi/4}^{pi/4}log[sqrt{2}]*[frac{1}{sqrt{2}}sinx+frac{1}{sqrt{2}}cosx]=int_{-pi/4}^{pi/4}log(sqrt{2}sin[x+pi/4])dx\end{array} \) Put x+π/4=t \(\begin{array}{l}I = int_{0}^{pi/2}log[sqrt{2}]sintdt\=int_{0}^{pi/2}log(sqrt{2}+log sintdt\=logsqrt{2}int_{0}^{pi/2}1dt+int_{0}^{pi/2}log sintdt\=pi/2 logsqrt{2}-pi/2 log2\=-pi/4 log2\end{array} \) Check... View Article

Find the value of x: 2x * 3 x+4 = 7x

log(2x)+log(3 x+4)=log(7x) x∗log(2)+(x+4)∗log(3)=x∗log(7) Simplifying a little bit more, x∗log(2)+x∗log(3)+4log(3)=x∗log(7) x(log(7)−log(2)−log(3))=4log(3) x=4log(3)log(7)−log(2)−log(3) We know that the values of log(2)=0.3010 log(3)=0.4771 log(7)=0.8451 On... View Article