View attachment 31635
Can someone please help me with part d, I've done all the other parts except for it? I'm not really sure how to approach it, but i know part c must be used somewhere in there.
Thanks
Using part c), here,
, so
.
Part c) tells you to replace every
x in the integrand with (
a +
b –
x), and that the integral will be the same value.
In this case, when you replace the numerator's
x with
, it becomes
=\sin^2 x)
. Notice that the denominator actually remains the same, because
)
becomes
\left ( \pi -2\left ( \frac{\pi}{2}-x \right ) \right ))
, which, if you do the algebra, is the same as
.
So by part c), we have
}$ d$x=\int_{\frac{\pi}{3}}^{\frac{\pi}{6}}\frac{\sin^2 x}{x(\pi - 2x)}$ d$x=\int_{\frac{\pi}{3}}^{\frac{\pi}{6}}\frac{1-\cos^2 x}{x(\pi - 2x)}$ d$x)
(by the Pythagorean trig. identity for the last equality).
Now, rearranging the equality
}$ d$x=\int_{\frac{\pi}{3}}^{\frac{\pi}{6}}\frac{1-\cos^2 x}{x(\pi - 2x)}$ d$x)
yields
}$ d$x=\int_{\frac{\pi}{3}}^{\frac{\pi}{6}}\frac{1}{x(\pi - 2x)}$ d$x=\frac{2}{\pi}\ln 2)
(by part b))
so that
}$ d$x=\frac{2}{\pi}\ln 2 \Rightarrow \int_{\frac{\pi}{3}}^{\frac{\pi}{6}}\frac{\cos^2 x}{x(\pi - 2x)}$ d$x=\frac{\ln 2}{\pi})
.
