# Math Help - Needs help on a tough question

1. ## Needs help on a tough question

Question: Use the fact that 7 cos x - 4 sin x = (3/2)(cos x + sin x) + (11/2)(cos x - sin x) to find the exact value of (upper limit: 0.5?, lower limit: 0)?((7 cos x - 4 sin x)/(cos x + sin x)) dx.

I need help with the steps, thanks in advance!

2. Hello, margaritas!

Edit: I got your correction . . . I'll edit my solution.

Use the fact that: $7\cos x - 4\sin x \:= \:\frac{3}{2}(\cos x + \sin x) + \frac{11}{2}(\cos x - \sin x)$

to find the exact value of: . $\int^{\frac{\pi}{2}}_0\frac{7\cos x - 4\sin x}{\cos x + \sin x}\,dx$

Answer: $\frac{3}{4}\pi$ . . . Right!

The function is: . $\frac{7\cos x - 4\sin x}{\cos x + \sin x}\;= \;\frac{\frac{3}{2}(\cos x + \sin x) + \frac{11}{2}(\cos x - \sin x)}{\cos x + \sin x}$

. . $= \;\frac{\frac{3}{2}(\cos x + \sin x)}{\cos x + \sin x} + \frac{\frac{11}{2}(\cos x - \sin x)}{\cos x + \sin x}$ $=\;\frac{3}{2} + \frac{11}{2}\cdot\frac{\cos x - \sin x}{\cos x + \sin x}$

The integration is: . $\frac{3}{2}\!\int\!\! dx + \frac{11}{2}\!\!\int\frac{\cos x - \sin x}{\cos x + \sin x}\,dx$

. . For the second integral, let $u = \cos x + \sin x\quad\Rightarrow\quad du = (\cos x - \sin x)\,dx$

. . then we have: . $\int\frac{du}{u} \:=\:\ln|u| \:=\:\ln|\cos x + \sin x|$

Hence, we have: . $\frac{3}{2}x + \frac{11}{2}\ln|\cos x + \sin x|\:\bigg]^{\frac{\pi}{2}}_0$

Evaluate: . $\left[\frac{3}{2}\left(\frac{\pi}{2}\right) + \frac{11}{2}\ln\left|\cos \frac{\pi}{2} + \sin \frac{\pi}{2}\right|\right] -$ $\left[\frac{3}{2}\cdot0 + \frac{11}{2}\ln|\cos0 + \sin0|\right]$

. . $= \;\left[\frac{3}{4}\pi + \frac{11}{2}\ln(0 + 1)\right] - \left[0 + \frac{11}{2}\ln(1 + 0)\right] \;= \;\boxed{\frac{3}{4}\pi}$

3. I recommend in this type of problem to use a Weierstrauss Substitution because it is a rational function of sine and cosine.

4. Oops, all the '?'s should be pi's instead.

But I think the solution by soroban will be helpful so thanks muchly!