Integration of sin cube x | Integral of sin^3x

The integration of sin cube x is equal to -cosx + cos³x/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 sin³x is given by

• ∫sin³x dx = (-3cos x)/4 + (cos 3x)/12 +C
• ∫sin³x dx = -cosx + cos³x/3+C.

What is the integral of sin^3x

Answer: The integral of sin^3x is equal to -cosx + cos³x/3+C.

Proof:

To find the integral of sin³x, we will use the following formula:

sin²x+cos²x=1.

⇒ sin²x=1-cos²x …(∗)

Now, ∫sin³x dx = ∫sin²x sinx dx

= ∫(1- cos²x) sinx dx by the above formula (∗)

= ∫sinx dx – ∫cos²x sinx dx

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

= -cosx + ∫z² dz +C as we know that ∫sinx dx=-cosx.

= -cosx + z³/3+C

= -cosx + cos³x/3+C

So the integral of sin^3x is equal to -cosx + cos³x/3+C where C is an integral constant.

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Integration of cube root x

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 – 4sin³x.

That is, 4sin³x = 3sinx -sin3x

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

Now integrating both sides, we will get that

∫sin³x 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.

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