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# Can you help me find the antiderivative of...? please...help?

f(x) = (sinx + cosx)^2

Im reviewing for a test and came across this question and just cant seem to get it!...help would be greatly appreciated...

### 4 Answers

- 9 years agoFavorite Answer
I think f(x) should be f'(x) in your question.

i.e. f'(x) = (sinx +cosx)^2

= sinx^2 +cosx^2 +2sinxcosx

= 1+ 2sinxcosx

= 1 + sin2x

Integrating

f(x) = x + 2cos2x+C

Source(s): Me - Shaun DizzleLv 79 years ago
Multiply it all out:

sin^2x + 2sinxcosx + cos^2x dx

Rearrange:

sin^2x + cos^2x + 2sinxcosx dx

1 + 2sinxcosx dx

Break it up into two integrals:

Unfortunately this one needs a u sub

â«1 + â«2sinxcosx dx

Only do u sub for right side:

u = sinx

du = cosx dx

â«1 + â«2 u du -----> the u we said is sinx, and the "du" sucked up the cosxdx

Now we have:

x + 2((u^2)/2) + C

= x + u^2 + C

Replace u with what we said it was..... remember we said u= sinx

x + sinx^2 + C

- ?Lv 49 years ago
f(x) = (sinx + cosx)^2

f(x) = sin^2(x) + 2sinxcosx + cos^2(x)

*sin^2(x) + cos^2(x) = 1

f(x) = 2sinxcosx + 1

*2sinxcosx = sin(2x)*

f(x) = 1 + sin(2x)

int(1+sin(2x)) = x + -(1/2)cos(2x) + C

antiderivative of f(x) = x - (1/2)cos(2x) + C

Without being arrogant, I'll just tell you that I am the only person who has yet given you the correct answer.

And to Sydney, if you integrate a derivative your just get the original function. There is no reason you can only find the derivative of f'(x)...

- 9 years ago
you would to take the integral of that which is integral ((sin(x)+cos(x))^2 dx

= integral (sin^2(x) + 2sin(x)*cos(x)+ cos^2(x)) dx

= -(1/2)cos(2x)+x