The integration of sin cube x is equal to -cosx + cos^{3}x/3+C where C denotes an integral constant. In this post, we will learn how to integrate sin^3x.

**Sine cube x integration formula:** The integration formula of sin^{3}x is given by

- ∫sin
^{3}x dx = (-3cos x)/4 + (cos 3x)/12 +C - ∫sin
^{3}x dx = -cosx + cos^{3}x/3+C.

## What is the integral of sin^3x

**Answer:** The integral of sin^3x is equal to -cosx + cos^{3}x/3+C.

*Proof:*

To find the integral of sin^{3}x, we will use the following formula:

sin^{2}x+cos^{2}x=1.

⇒ sin^{2}x=1-cos^{2}x …(∗)

Now, ∫sin^{3}x dx = ∫sin^{2}x sinx dx

= ∫(1- cos^{2}x) sinx dx by the above formula (∗)

= ∫sinx dx – ∫cos^{2}x sinx dx

[Let cosx =z, so we have -sinx dx =dz]

= -cosx + ∫z^{2} dz +C as we know that ∫sinx dx=-cosx.

= -cosx + z^{3}/3+C

= -cosx + cos^{3}x/3+C

So the integral of sin^3x is equal to -cosx + cos^{3}x/3+C where C is an integral constant.

**ALSO READ:**

Let us now find the integral of sine cube x using the formula of sin3x.

## Find the Integral of sin cube x

**Answer:** The integral of sin cube x is equal to (-3cos x)/4 + (cos 3x)/12 +C.

*Proof:*

Let us recall the sin3x formula: sin3x=3sinx – 4sin^{3}x.

That is, 4sin^{3}x = 3sinx -sin3x

∴ sin^{3}x = $\dfrac{1}{4}$ × (3sinx -sin3x)

Now integrating both sides, we will get that

∫sin^{3}x dx = ∫$\dfrac{1}{4}$ × (3sinx -sin3x) dx

= ∫$\dfrac{3}{4}$ sinx dx – ∫$\dfrac{1}{4}$ sin3x dx

= $\dfrac{3}{4}$∫sinx dx – $\dfrac{1}{4}$∫sin3x dx

= $\dfrac{3}{4}$(-cosx) – $\dfrac{1}{4}$ $\dfrac{-\cos 3x}{3}$ +C, obtained by the formula ∫sinmx dx = $-\dfrac{\cos mx}{m}$

= $\dfrac{-3 \cos x}{4}$ + $\dfrac{\cos 3x}{12}$ +C

So the integration of sine cube x is equal to (-3cos x)/4 + (cos 3x)/12 +C where C is an integration constant.

Have You Read These Integrals?

## FAQs

**Q1: What is the integration of sin^3x?**

Answer: The integration of sin^3x is (-3cos x)/4 + (cos 3x)/12 +C, C denotes an arbitrary constant.

**Q2: What is the integration formula of sin cube x?**

Answer: The integration of sin^3x formula is as follows: ∫sin^{3}x dx = (-3cos x)/4 + (cos 3x)/12 +C. Here C is an arbitrary integral constant.