Find constants A and B

**George321** · May 3rd 2010, 08:12 AM

Given that
$A(sin\theta+cos\theta)+B(cos\theta-sin\theta)=4sin\theta$
Find the constants $A$ and $B$ .

Can't see what i need to do here.
Thanks

**Soroban** · May 3rd 2010, 08:50 AM

Hello, George321!

Given that: . $A(\sin\theta+\cos\theta)+B(\cos\theta-\sin\theta)\:=\:4\sin\theta$

Find the constants $A$ and $B$

The left side is: . $A\sin\theta + A\cos\theta + B\cos\theta - B\sin\theta \;=\;(A-B)\sin\theta + (A+B)\cos\theta$

The equaton becomes: . $(A-B)\sin\theta + (A+B)\cos\theta \;=\;4\sin\theta$

Equate coefficients: . $\begin{Bmatrix}A - B &=& 4 \\ A+B &=& 0 \end{Bmatrix}$ . . . . and solve the system.

**George321** · May 3rd 2010, 11:58 AM

Thanks for your help, im now stuck on the second bit though...
$<br /> \int_{0}^{\frac{1}{4}\pi}\frac{4sin\theta}{sin\the ta+cos\theta}d\theta$
giving the answer in the form $a\pi-lnb$

Can't remember how to integrate this kind of equation...

**TheCoffeeMachine** · May 5th 2010, 01:16 AM

$\int_{0}^{\frac{\pi}{4}}{\dfrac{4\sin{\theta}}{\si n{\theta}+\cos{\theta}}\;{d\theta}}$ $= \int_{0}^{\frac{\pi}{4}}{\dfrac{2\left(\sin{\theta }+\cos{\theta}\right)-2\left(\cos{\theta}-\sin{\theta}\right)}{\sin{\theta}+\cos{\theta}}\;{ d\theta}}$ $= \int_{0}^{\frac{\pi}{4}}\left(\dfrac{2\left(\sin{\ theta}+\cos{\theta}\right)}{\sin{\theta}+\cos{\the ta}}-\dfrac{2\left(\cos{\theta}-\sin{\theta}\right)}{\sin{\theta}+\cos{\theta}}\ri ght)\;{d\theta}$ $= \int_{0}^{\frac{\pi}{4}}\left(2-\dfrac{2\left(\cos{\theta}-\sin{\theta}\right)}{\sin{\theta}+\cos{\theta}}\ri ght)\;{d\theta}$ $= \int_{0}^{\frac{\pi}{4}}\left(2-2\left\{\dfrac{\cos{\theta}-\sin{\theta}}{\sin{\theta}+\cos{\theta}}\right\}\r ight)\;{d\theta}$ $= \bigg[2\theta-2\log\left(\sin{\theta}+\cos{\theta}\right)\bigg]_{0}^{\frac{\pi}{4}} = \boxed{\dfrac{1}{2}\pi-\ln(2)}.$

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