Question

i have a problem this integral. i keep making a mistake but i can't see it ∫−sin2t+sint−sin3tdt

Answer 1

\[\int\limits_{0}^{\pi/2} -\sin^2t+sint-\sin^3t dt\]

Answer 2

Oh, those arepowers, gotcha.

Answer 3

\[\int\limits_{0}^{\pi/2} -\sin^2t dt +\int\limits_{0}^{\pi/2} sint dt - \int\limits_{0}^{\pi/2} \sin^3t dt\]

Answer 4

for the first integral i got \[\frac{ -\pi }{ 4 }\]

Answer 5

Im working the whole problem at a time, so I havent gotten to that. Ill work just that integral

Answer 6

for the second one -1 and for the third one \[\frac{ 2 }{ 3 }\]

Answer 7

The -pi/4 looks right, but I usually don't apply the limits individually. Lemme finish the problem and see what I get :3

Answer 8

the problem is that in the end a get both fractions with minus. and in the book only the one with pi has minus

Answer 9

Hmm, I got the same thing first go around.

Answer 10

But yes, it appears we did make a sign mistake, I checked.

Answer 11

Found where I did the stupid sign mistake.

Answer 12

where?

Answer 13

On the sin^3 part.
\[\int\limits_{}^{}-\sin ^{3}xdx\]
\[-\int\limits_{}^{}(1-\cos ^{2}x)(sinx)dx\]
\[-(\int\limits_{}^{}sinxdx-\int\limits_{}^{}sinxcos ^{2}xdx)\]
u = cosx, du = -sinxdx, dx = -du/sinx
\[-(\int\limits_{}^{}sinxdx+\int\limits_{}^{}u ^{2}du)\]
\[-(-cosx+\frac{ \cos ^{3}x }{ 3 })\]
Now of course the cosx we know cancels partwat through, so we end up with
\[-\frac{ \cos ^{3}x }{ 3 }\]from 0 to pi/2
\[-0-(-\frac{ 1 }{ 3 })=\frac{ 1 }{ 3 }\]
So it does end up being positive.

Answer 14

thanks

Answer 15

Np :3