1. ## definite integrate

evaluate $\int^{\frac{\pi}{3}}_0(sinx-cosx)^2 dx$

what i did

$\int^{\frac{\pi}{3}}_0(sinx-cosx)^2 dx=[\frac{(sinx-cosx)^3}{3(-cosx-sinx)}]^{\frac{\pi}{3}}_0$

is this right?

2. I think you can treat it as:

$\int (f(x))^n dx = \frac{f^{(n+1)}(x)}{(n+1)(f'(x))}$

EDIT: After you edited your post, yes, that's what I would have done

3. i posted what i did using the formula u posted, mind to check if i did it right

4. Hm... if y = sin(x)

Then dy/dx becomes cos(x)

And if y = cos(x),

Then dy/dx = -sin(x)

What I mean is that you put the differential of the function and not the integral of the function in the denominator.

5. the 2 posts u posted are contradicting.. so am i right?

6. Sorry, I have edited my first post without actually taking a look at whether the integral was right or wrong.

My second post says it.

In fact, it's only my edit in my first post which is contradicting. The 'formula' that I posted remains the same and applies.

So, your integral is wrong. You need to put the differential of the function as denominator, not its integral.

$y = (x^2 + x + 1)^3$

This becomes:

$\frac{dy}{dx} = (x^2 + x + 1)^2 . (2x + 1)$

When you integrate, you divide instead of multiplying the differential of the function.

7. i dont know what went wrong but substituting the values in doesnt give me the correct answer...would u work it out and tell me what u get?

8. $\int^{\frac{\pi}{3}}_0(sinx-cosx)^2 dx = [\frac{(sinx - cosx)^3}{3(cosx + sinx)}]^{\frac{\pi}{3}} _0$

$= \frac{(sin(\frac{\pi}{3}) - cos(\frac{\pi}{3}))^3}{3(cos(\frac{\pi}{3}) + sin(\frac{\pi}{3}))} - \frac{(sin(0) - cos(0))^3}{3(cos(0) + sin(0))}$

$= \frac{(\frac{\sqrt{3}}{2} - \frac{1}{2})^3}{3(\frac12 + \frac{\sqrt{3}}{2})} - \frac{(0 - 1)^3}{3(1 + 0)}$

$= \frac{(\frac{\sqrt{3}-1}{2})^3}{3(\frac{1+\sqrt{3}}{2})} - \frac{-1}{3}$

$= \frac{(\frac{(\sqrt{3}-1)^3}{8})}{\frac{3+3\sqrt{3}}{2}} + \frac{1}{3}$

$= \frac{6\sqrt{3}-10}{12 + 12\sqrt{3}} + \frac{1}{3}$

If I didn't make any mistake, you can now further simplify.

EDIT: Seems like it's wrong.

9. Maybe taking another approach to have less of those square roots...

ok, I'll try this.

$\int^{\frac{\pi}{3}}_0 (sin(x) - cos(x))^2 dx = \int^{\frac{\pi}{3}}_0 sin^2(x) - 2sin(x)cos(x) + cos^2(x) dx$

$= \int^{\frac{\pi}{3}}_0 1 - sin(2x)dx$

$= [x+ \frac12 cos(2x)]^{\frac{\pi}{3}}_0$

$= [\frac{\pi}{3} + \frac12 cos(\frac{2\pi}{3})] - [ 0 + \frac12 cos(0)]$

$= \frac{\pi}{3} -\frac14 - \frac12$

$= \frac{\pi}{3} -\frac34$

That's better

10. i see my mistake, i divided by the integral!!!!

11. Originally Posted by Unknown008
$\int^{\frac{\pi}{3}}_0(sinx-cosx)^2 dx = [\frac{(sinx - cosx)^3}{3(cosx + sinx)}]^{\frac{\pi}{3}} _0$

$= \frac{(sin(\frac{\pi}{3}) - cos(\frac{\pi}{3}))^3}{3(cos(\frac{\pi}{3}) + sin(\frac{\pi}{3}))} - \frac{(sin(0) - cos(0))^3}{3(cos(0) + sin(0))}$

$= \frac{(\frac{\sqrt{3}}{2} - \frac{1}{2})^3}{3(\frac12 + \frac{\sqrt{3}}{2})} - \frac{(0 - 1)^3}{3(1 + 0)}$

$= \frac{(\frac{\sqrt{3}-1}{2})^3}{3(\frac{1+\sqrt{3}}{2})} - \frac{-1}{3}$
this method probably cannot be used as the answer is wrong!!

yr 2nd method is right

any experts can clarify if it is wrong?

12. http://www.wolframalpha.com/input/?i=\int^{pi%2F3}_{0}+(sin(x)+-+cos(x))^2

I just checked. The latter solution is supposed to be the correct one.

So, I guess the first method is not the correct one