Question

Using substitution formula to evaluate the integral ^pi/4[0 (1-cos2t)sin2t dt?

Answer 1

\[\int\limits_{0}^{\frac{ \pi }{ 4 }} (1-\cos2t)\sin2t dt\]

Answer 2

so did you try substituting
u = 1-cos 2t
?

Answer 3

so far I have u=1-cos2t) and du=sin2t dt
Can someone tell me if I have this substation done correctly?

From there I have t=0; 1-cos2(0)=0
t(pi/4)=1-cos2(pi/4)= and I seem to be stuck right here if my above equation is correct.

Answer 4

correct substitution :)

Answer 5

cos 2(pi/4) = cos pi/2 = ... ?

Answer 6

Oh I see I did not cross cancel the two from pi/4 . Thank you.

Now it should be \[\int\limits_{0}^{\pi/4}udu=\int\limits_{0}^{\pi/4} u(\sin2t) \frac{ du }{ 2\sin2t }\] now the (sin2t) crosses out with the fraction of du/2 sin2t leaving 1/2.

Now I did get done talking to my teacher about this problem and she said the problem should look like this \[\int\limits_{0}^{\pi/4} \frac{ 1 }{ 2}*\frac{ u^2 }{ 2 }\]

can someone explain where the u^2/2 came from?

Answer 7

the problem should look like
\(\Large \int \limits _0^1 (u/2) du\)

Answer 8

while doing substitution, we change the limits also!

Answer 9

so,
u = 1-cos 2t
0--->0
pi/4 --->1

du/2 = sin 2t dt

clear till here ?
substitute all these in you integral to form a new easier integral