How do I compute the integral $\int_0^{\pi/2} \cfrac 1 {12\cos x + 9 \sin x} dx?$
Please can you help.
How do I compute the integral $\int_0^{\pi/2} \cfrac 1 {12\cos x + 9 \sin x} dx?$
Please can you help.
HINT
Rewrite $12 \cos(x) + 9 \sin(x)$ as $15 \left(\dfrac{12}{15} \cos(x) + \dfrac{9}{15} \sin(x) \right) = 15 \cos(x + \theta)$ where $\tan(\theta) = - \dfrac9{12}$. (Note that $15 = \sqrt{9^2 + 12^2}$.) This is similar to the problem here.
This is a solution from WolframAlpha, I still don't know if this kind of thing is allowed here on MSE.
The solution is indefinite, you can compute the definite integral after.
If all else fails, the Weierstrass substitution will do it.
However, let's try something that begins with what Marvin suggested in his answer. First notice that $12^2+9^2=15^2$, so $\left(\frac{12}{15}\right)^2+\left(\frac{9}{15}\right)^2=1$. Hence we can find $\theta$ such that $\cos\theta=12/15$ and $\sin\theta=9/15$. Then $ 12\cos x + 9\sin x = 15\left(\frac{12}{15}\cos x+\frac{9}{15}\sin x\right) = 15(\cos\theta\cos x + \sin\theta\sin x) = 15\cos(x-\theta). $ Therefore $ \int\frac{dx}{12\cos x+9\sin x} = \int \frac{dx}{15\cos(x-\theta)}= \int \frac{\cos(x-\theta)\,dx}{15(\cos(x-\theta))^2} = \frac{1}{15}\int \frac{\cos(x-\theta)\,dx}{1- (\sin(x-\theta))^2} $ $ = \frac{1}{15}\int \frac{du}{1-u^2} = \frac{1}{15}\int \frac{A}{1-u} + \frac{B}{1+u} \, du. $ Then use partial fractions.