Question

Use integration by parts to evaluate the indefinite integral
[ 2x sin(c) dx

Answer 1

-2x cos x + 2 sin x + C

Explanation:

∫ (2x sin x) dx

If u = 2x, then du = 2 dx.

If dv = sin dx, then v = -cos x.

∫ u dv = uv − ∫ v du

= 2x (-cos x) − ∫ (-cos x) (2 dx)

= -2x cos x + 2 ∫ cos x dx

= -2x cos x + 2 sin x + C

Answer 2

`\displaystyle \large{2 \sin x - 2x \cos x + C}`

Explanation:

We are given the indefinite integral:

`\displaystyle \large{\int {2x \sin x} \ dx }`

Using by-part method, we have to substitute u-term and dv appropriately.

By-part is an integration of product rules, when integrated the product rules of differentiation, we’ll obtain:

`\displaystyle \large{\int {u} \ dv = uv - \int {v} \ du}`

Above is by-part method/formula.

Where 4 terms are presented:

• u
• du
• dv
• v

Our main terms to substitute are u and dv which mean u-term has to be a function that’s differentiatable and dv has to be a function that’s integratable.

The main concept of by-part is to understand how to substitute appropriately which you can simply follow below:

LIATE

Stands for Logarithm, Inverse (Trigonometry), Algebraic, Trigonometric and Exponential.

These are in orders from first to last on what to let u-term first. That means logarithm functions must be the first to substitute themselves as u-term, so if you encounter a logarithmic function and a polynomial function, you must let u = logarithmic function while dv = polynomial.

In this case, we have 2x which is polynomial and sin(x) which is trigonometric. According to LIATE, we have to let Algebraic or Polynomial 2x be first to substitute as u-term, that means our dv is trigonometric sin(x).

Therefore, we have:

• u = 2x
• du = 2dx
• dv = sin(x)dx
• v = -cos(x)

Now, substitute these terms in accordingly to formula of by-part.

`\displaystyle \large{\int {2x \sin x} \ dx = 2x \cdot (-\cos x) - \int {-\cos x \cdot 2 \ dx}}\\ \displaystyle \large{\int {2x \sin x} \ dx = -2x \cos x + \int {2 \cos x \ dx}}\\ \displaystyle \large{\int {2x \sin x} \ dx = -2x \cos x + 2 \int {\cos x \ dx}}\\ \displaystyle \large{\int {2x \sin x} \ dx = -2x \cos x + 2 \cdot \sin x + C}\\ \displaystyle \large{\int {2x \sin x} \ dx = -2x \cos x + 2 \sin x + C}`

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Summary

Property

`\displaystyle \large{ \int {k f(x)} \ dx = k \int{f(x)} \ dx \ \ \ \ \tt{(k \ \ \ is \ \ \ a \ \ \ constant.)}\\`

Only shown in explanation.

By-Part

`\displaystyle \large{\int u \ dv = uv- \int v \ du}`

LIATE

The functions in order that should be u-term from first to last.

• Logarithm
• Inverse Trigonometric

These functions above do not have integration formula by default.

• Polynomial (Algebraic)
• Trigonometric
• Exponential (Last since it’s the easiest to integrate, especially natural exponential)

Indefinite Integral

Make sure to always add + C after evaluating the integral, regardless what multiplies or attempts to affect + C, we must always add + C.

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