# Intergral of sin^3(x)cos(x)+cos(x)

• mcelgiraffe
In summary, the two expressions are equivalent when evaluated indefinite integrals, but they may not be equal when evaluated as definite integrals.
mcelgiraffe
Hi everyone,

I have worked the following integral two different ways and received two different answers.

∫sin^3 xcosx dx+ ∫cosx dx

The first method
= ∫sin^3 xcosx dx+ ∫cosx dx
Let u = sinx
du/dx = cosx
dx = 1/cosx du
= ∫u^3 cosx*(1/cosx) du + ∫cosx dx
= ∫u^3 du + ∫cosx dx
= (1/4) u^4+ sinx+C
= (1/4)sin^4 x+sinx+C

The second method
= ∫sinx(sin^2x)cosx dx + ∫cosxdx
= ∫sinx(1-cos^2x)cosx dx + ∫cosx dx
Let u = cosx
du/dx = -sinx
dx = (-1/sinx) du
= ∫sinx(1-u^2 )u* (-1/sinx) du + ∫cosx dx
= -∫(u-u^3)du+∫cosx dx
= (-1/2)u^2+ (1/4)u^4 + sinx + C
= (-1/2)cos^2x + (1/4)cos^4x + sinx + C

If I graph the two answers they are the same graph, but one is shifted vertically by 1/2. Now as an indefinite integral it seems as if this may only be different values of C. But if this were a definite integral wouldn't we drop the C? Meaning that these two equations are not equal? Have I done something wrong in the second equation?

Thank You, James

mcelgiraffe said:
If I graph the two answers they are the same graph, but one is shifted vertically by 1/2. Now as an indefinite integral it seems as if this may only be different values of C. But if this were a definite integral wouldn't we drop the C?
You're missing a key idea, either about what C represents, or about the relationship between definite integrals and antidifferentiation.

But no matter -- that can be fixed by experimentation! Pick your favorite value for C, and compute the definite integral from 0 to 1. Now pick another value for C, and compute again...

(approximate values are fine!)

mcelgiraffe said:
Hi everyone,

I have worked the following integral two different ways and received two different answers.

∫sin^3 xcosx dx+ ∫cosx dx

The first method
= ∫sin^3 xcosx dx+ ∫cosx dx
Let u = sinx
du/dx = cosx
dx = 1/cosx du
= ∫u^3 cosx*(1/cosx) du + ∫cosx dx
= ∫u^3 du + ∫cosx dx
= (1/4) u^4+ sinx+C
= (1/4)sin^4 x+sinx+C

The second method
= ∫sinx(sin^2x)cosx dx + ∫cosxdx
= ∫sinx(1-cos^2x)cosx dx + ∫cosx dx
Let u = cosx
du/dx = -sinx
dx = (-1/sinx) du
= ∫sinx(1-u^2 )u* (-1/sinx) du + ∫cosx dx
= -∫(u-u^3)du+∫cosx dx
= (-1/2)u^2+ (1/4)u^4 + sinx + C
= (-1/2)cos^2x + (1/4)cos^4x + sinx + C

If I graph the two answers they are the same graph, but one is shifted vertically by 1/2. Now as an indefinite integral it seems as if this may only be different values of C. But if this were a definite integral wouldn't we drop the C? Meaning that these two equations are not equal? Have I done something wrong in the second equation?

Thank You, James

They are indeed equivalent. If you would have a definite integral, the final result would also be the same because the shift between the two expressions would cancel out. You can check with some numbers and see that it works out.

Thank You both for your response. My instructor told me the same thing but I had not accepted it yet. I did work it out both ways from 0 to pi and got the same answers. There is still something that I am uneasy about. I will spend some more time with it and see if I can get my question more organized. Thanks again!

## 1. What is the formula for the integral of sin^3(x)cos(x)?

The formula for the integral of sin^3(x)cos(x) is -1/4 cos^4(x) + C, where C is the constant of integration.

## 2. How do you solve the integral of sin^3(x)cos(x)?

To solve the integral of sin^3(x)cos(x), you can use the trigonometric identity sin^2(x) = (1-cos(2x))/2 to rewrite the expression as (1-cos(2x))sin(x)cos(x). Then, you can use the substitution method to solve for the integral.

## 3. Can you use the power rule to solve the integral of sin^3(x)cos(x)?

No, the power rule cannot be directly applied to solve the integral of sin^3(x)cos(x) because the power rule only applies to integrals of the form x^n, where n is a constant.

## 4. What is the relationship between the integral of sin^3(x)cos(x) and the integral of cos(x)?

The integral of sin^3(x)cos(x) is a more complex function than the integral of cos(x), which is simply sin(x) + C. However, both integrals can be solved using similar techniques, such as substitution and trigonometric identities.

## 5. Can the integral of sin^3(x)cos(x) be solved using integration by parts?

Yes, the integral of sin^3(x)cos(x) can be solved using integration by parts. However, it may be more straightforward to use the substitution method instead.

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