Solve the Integral: \int \sin(2x).\sin(x)dx | Expert Help Available

[tiagobt]

Could anyone help me solve the following integral?

$$\int \sin(2x).\sin(x)dx$$

Thanks

 

[arildno]

Use the product/sum formula for trigonometric functions.

 

[tiagobt]

Use the product/sum formula for trigonometric functions.

Do you mean $$sin(2x)=2.sinx.cosx$$ ? Like this:

$$\int 2sinx.cosx.sinx.dx$$
$$= 2 \int sin^2x.cosx.dx$$
$$= 2 \int (1 - cos^2x)cosx.dx$$
$$= 2 \int cosx dx - 2\int cos^3x.dx$$

Is that what you mean? I can solve the first integral... But not the second.

 

[arildno]

It wasn't what I meant, but since your own procedure can be used as well, I'll help you out along the track you chose.
You have established:
$$\int\sin(2x)\sin(x)dx=2\int\sin^{2}x\cos(x)dx$$
Now, use the substitution
$$u=\sin(x)$$
Then, we have:
$$\frac{du}{dx}=\cos(x)\to{dx}=\frac{du}{\cos(x)}$$
Thus, we have gained:
$$\int\sin(2x)\sin(x)dx=2\int{u}^{2}du=\frac{2}{3}u^{3}+C=\frac{2}{3}\sin^{3}x+C$$

 

[tiagobt]

Thanks! It's actually easier than I thought.

 

[arildno]

If you are interested in another way to do this, we have for any choices a,b the equalities:
$$\cos(a+b)=\cos(a)\cos(b)-\sin(a)\sin(b),\cos(a-b)=\cos(a)\cos(b)+\sin(a)\sin(b)$$
Combining these, we gain:
$$\sin(a)\sin(b)=\frac{1}{2}(\cos(a-b)-\cos(a+b))$$
That is,
$$\sin(2x)\sin(x)=\frac{1}{2}(\cos(x)-\cos(3x))$$

This is what I meant with "using the product/sum formula".