Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Find the area between $y=9\sin x\,$ and $\,y = 10\cos x\,$ over the interval $[0, \pi]$.

Having a tough time with this problem. Im good with finding the area between two curves but the sin and cos threw me off a lot.

share|cite|improve this question
Have you tried graphing or drawing it to get an idea of what it looks like? – ferson2020 Feb 6 '13 at 20:22
okay thank you. sorry – Ak47 Feb 7 '13 at 0:01
up vote 2 down vote accepted

You'll want to find the point of intersection, and use the $t=x$-value of this point as a bound for integrating (see note below).

Set the two equations equal to one another, and solve for the values $t$ (the x-coordinate) that solves the resulting equation: there will be one such value on your domain.

$$9\sin(t) = 10\cos(t)\implies \frac{\sin t}{\cos t} = \frac {10}{9}\implies \tan t = \frac{10}{9} \implies t = \tan^{-1} \frac{10}{9}$$

Then calculate the sum of the integrals $$\int_0^t (10\cos x - 9\sin x)\,dx + \int_t^\pi (9\sin x- 10\cos x)\,dx.$$

Note: you need to know the point of intersection to divide the integral into two parts, because from $0\leq x \lt t$, where t is the value of x at the point of intersection, $10\cos x > 9\sin x$. At $x = t$, $10\cos x = 9 \sin x$. And from $t \lt x \leq \pi$, $9\sin x > 10 \cos x$. So to measure total area, we need to ensure we choose the correct function as the "upper bound" of each region when we integrate to obtain the total area of the regions bound by the functions.

enter image description here

share|cite|improve this answer
Great. You shouldn't call $x$ $x$, though, since it is also your integration variable. – 1015 Feb 6 '13 at 20:59
thanks @julien, good point. – amWhy Feb 6 '13 at 21:02

Hint: the curves look like this:

enter image description here

You need to find the point of intersection and split the interval into two pieces.

share|cite|improve this answer

As mrf suggested. Find the $x$ value at the intersection of the two curves. Then calculate the integral $\int_0^x(upper curve) - (lower curve)\,\mathrm{d}x$ + $\int_x^\pi(upper curve) - (lower curve)\,\mathrm{d}x$.

$$\int_0^x(10 \cos(x) - 9\sin(x))\,\mathrm{d}x + \int_x^\pi(9\sin(x) - 10 \cos(x))\,\mathrm{d}x$$

You should end up with $2\sqrt{181}$which is approximately 26.907.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.