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How do you find the integral of #int (1 + cos x)^2 dx#?

Answer 1

Aurora Alvarado

#int(1+cosx)^2dx=1/4(6x+8sinx+sin2x)+"c"#

First, apply the perfect square formula to expand the integrand.

#int(1+cos^2x) dx=int1+2cosx+cos^2xdx#

Then use the reduction formula on #cos^2x# to get it into an integrable form

#int1+2cosx+cos^2xdx=int1+2cosx+1/2+1/2cos2xdx=int3/2+2cosx+1/2cos2xdx#

Use the power rule or standard integrals to integrate each term.

#int3/2+2cosx+1/2cos2x=3/2x+2sinx+1/4sin2x+"c"=1/4(6x+8sinx+sin2x)+"c"#

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Answer 2

Luke Alvarez

#int(1+cos x)^2 dx# =#int(1+2cos x +cos^2 x) dx# =#int(1+2cos x +(1/2cos 2x - 1/2) dx# * =#int(1/2 + 2cos x + 1/2cos 2x) dx# =#1/2x +2sin x +1/4sin 2x +c#

*#1 + 2cos^2 x = cos 2x# #:. cos^2 x = 1/2cos 2x -1/2#

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Answer 3

Samantha Adkison

To find the integral of ∫(1 + cos x)^2 dx, you can expand the expression (1 + cos x)^2, which yields 1 + 2cos(x) + cos^2(x). Then, you can use trigonometric identities to rewrite cos^2(x) in terms of 1 + cos(2x)/2. Substituting these expressions back into the integral, you will have a polynomial in terms of cos(x). You can then integrate each term individually, which will involve straightforward power rule integration and trigonometric substitutions. Finally, you can simplify the result to obtain the integral of the original expression.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.