## anonymous 5 years ago integrate from 0 to pi/2 (sinx + cosx)/(9+16sin2x) dx

1. watchmath

Is that $$\sin(2x)$$ or $$\sin^2(x)$$?

2. anonymous

former

3. anonymous

you workin on it?

4. watchmath

This is very tricky. Notice that $$[4(\sin(x)-\cos(x))]^2=16(\sin^2(x)+\cos^2(x)-\sin(2x))=16-16\sin(2x)$$. So the denominator of the integreal is equal to $$25-(4(\sin x - \cos x))^2=[5-(4\sin x -4\cos x )]\cdot [5+(4\sin x-4\cos x)]$$ Now let $$u=4\sin x-4\cos x$$. Then $$du=4(\sin x+\cos x)\,dx$$. So the integral is equivalent to $\frac{1}{4}\int \frac{du}{(5-u)(5+u)}$ Then you may continue from there :).

5. anonymous

6. watchmath

$$\frac{1}{(5-u)(5+u)}=\frac{1}{10}(\frac{1}{5-u}+\frac{1}{5+u})$$

7. anonymous

how about $du/5^{2}-u ^{2}=1/10 \log(u+5)/(u-5)$

8. watchmath

yes, that is correct!

9. anonymous

i cant work it through... asked some other guys but had no luck

10. watchmath

well then your answer is $$\frac{1}{40}\ln|\frac{4\sin x-4\cos x+5}{4\sin x-4\cos x-5}|$$ and then you plug in $$x=\pi/2$$ and $$x=0$$.

11. anonymous

my book says the answer is 1/10 log3

12. watchmath

$$\frac{1}{40}(\ln(9)-\ln(1/9))=\frac{1}{40}(4\ln(3))=\frac{1}{10}\ln 3$$

13. anonymous

just a little simplification would hav done it.. thanx a lot mate

14. anonymous

Seriously, watchmath, where did you even think to rewrite it like that? I understand how to do the partial fractions/whatever. But I would never think to come up with something like the to rewrite it o.o

15. anonymous

he has got a pretty superior brain i guess...lol