Question

help please
(do not give irrelevant answers plz)​

Answer 1

Answer:

0.046 (you can approximate with log table, I used calculator)

Step-by-step explanation:

Let's first find the integral:

$\int {\frac{sin+cos}{16+9sin2x}} \, dx$

Let $u = sinx - cosx$

$du = (cosx + sinx)dx$

$u^2 = sin^2x+cos^x-2sinxcosx = 1-2sinxcosx = 1-sin2x$

$sin2x = 1-u^2$

$\int {\frac{1}{16+9(1-u^2)}} \, du$

$\int {\frac{1}{16+9-9u^2)}} \, du$

$\int {\frac{1}{25-9u^2}} \, du$

$\int {\frac{1}{5-3u}\cdot\frac{1}{5+3u}} \, du$

Using partial fraction decomposition:

$\int {\frac{1}{10}(\frac{1}{5-3u}+\frac{1}{5+3u})} \, du$

$\frac{1}{10}(\int {\frac{1}{5-3u}} \, du+\int {\frac{1}{5+3u}} \, du)$

$\frac{1}{10}(\frac{\ln{\mid5-3u\mid}}{-3} + \frac{\ln{\mid5+3u\mid}}{3})$

$\frac{1}{30}(\ln{\mid5+3u\mid} - \ln{\mid5-3u\mid})$

$\frac{1}{30}(\ln{\mid5+3(sinx-cosx\mid} - \ln{\mid5-3(sinx-cosx)\mid}$

Evaluating from 0 to $\frac{\pi}{4}$

$= 0 - 0.04620981203 = 0.04620981203 = 0.046$