Question

Integral.

Answer 1

my answer is 1/3 sin^3x - 2 cosx + 1/3 cos^3x + c.. am i right

Answer 2

what i did was expand the sinx + cosx and.. i got sin^2x + 2 sinxcosx + cos^2x dx

Answer 3

or actually, x -1/2 cos(2x) + c

Answer 4

you can simplify to sin^2x but it's harder...
well here's how I do it

Answer 5

thank you so much =)

Answer 6

\[\int\limits_{}^{} (\sin(x) + \cos(x))^{2} dx\]
Like you said, we can distribute it
\[(\sin(x) + \cos(x))^2 = \sin^2x + 2sinxcosx + \cos^2x\]

You know that sin^2x + cos^2x =1, so
\[(\sin(x) + \cos(x))^2 = 1 + 2sinxcosx\]

Then if you use the Double Angle identity, sin2x= 2sinxcosx, your equation is now
\[(\sin(x) + \cos(x))^2 = 1 \sin(2x)\]

So now, the integral becomes
\[\int\limits_{}^{} 1+ \sin(2x) dx\]

Answer 7

1+sin(2x) ***

Answer 8

ah yes yes yes. you are right

Answer 9

Then separate those integrals
\[\int\limits_{}^{}1 dx + \int\limits_{}^{} \sin(2x) dx\]
\[x - \frac{ 1 }{ 2 } \cos(2x) + C\]

Answer 10

integral of sin(2x) was in a table that I memorized from textbook.
If you're curious of how it's integrated step by step, you can look up "integration of sin(2x)"

Answer 11

Thank you soooo muuuuuuuuuch!! =)

Answer 12

You're welcomed :)