Evaluate $\int{(\sin x + 2\cos x)}^3 \, dx$

Question

I tried to expand the binomial. But its complicated.

$\int{\sin x}^3 + {(2\cos x)}^3 + 3\sin x2\cos x (\sin x + 2\cos x)\, dx$

I dont think i can use integral by parts on it.

Using substitution setting

$(\sin x + 2\cos x)=u$ i dont know how to continue..

Answer 1

Expansion is your best approach. Now use linearity:

$\int{\sin^3(x)} dx + 6\int{\cos(x)\sin^2(x)} dx + 12\int{\cos^2(x)\sin(x)} dx + 8\int{\cos^3(x)} dx$

For the first and fourth integrals, rewrite them using $\sin^3(x) = (1-\cos^2(x))\sin(x)$ and $\cos^3(x) = (1-\sin^2(x))\cos(x)$. Now use substitution for all four integrals.

Answer 2

Each of the resulting integrals from expanding the binomial are not particularly complicated.

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For $\int \sin^3(x)dx$, you can note the power reducing formula

$$\int \sin^m(x)dx = - \frac{\sin^{m-1}(x) \cos(x)}{m} + \frac{m-1}{m} \int \sin^{m-2}(x)dx$$

That $m=3$ in this case makes it easy.

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The integral $\int \cos^3(x)dx$ has a similar power-reduction formula to use:

$$\int \cos^m(x)dx = \frac{\cos^{m-1}(x) \sin(x)}{m} + \frac{m-1}{m} \int \cos^{m-2}(x)dx$$

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For $\int \sin^2 x \cos x dx$, make the $u$-substituion $u=\sin(x)$, which gives $du = \cos(x)dx$. Then you can just apply the power rule.

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For $\int \sin(x)\cos^2(x)dx$, similarly, just let $u = \cos(x)$. Then $du = -\sin(x)dx$.

Answer 3

You used $(a+b)^3 = a^3 + 3a^2 b + 3ab^2 + b^3$ wrong. It should be $$\int \sin^3 x + 3 \sin^2 x \cdot 2 \cos x + 3 \sin x \cdot 4 \cos^2 x + 8 \cos^3 x \, dx $$ You can integrate $$\int \sin^3 x \, dx = \int(1-\cos^2 x)\sin x \, dx = -\int \frac{d}{dx}\left(\cos x - \frac13\cos^3 x \right) dx = -\cos x + \frac13\cos^3 x+C$$ The strategy:

1. take out one copy of $\sin$ or $\cos$. In the above it was $-\sin x$. This is because $\frac{d}{dx}\cos x = - \sin x$.
2. Identify a polynomial in the other trig function. In the above, we found $ 1 - u^2$ where $u=\cos x$.
3. Now use chain rule "in reverse" to identify the integrand as a derivative. In the above, we noted that $\int 1 - u^2 du = u - \frac13u^3 $.
4. Conclude.

See if you can do the rest with a similar method.

Answer 4

$$\int{(\sin x + 2\cos x)}^3dx=$$ $$=\int(\sin^3x+6\sin^2x\cos{x}+12\sin{x}\cos^2x+8\cos^3x)dx=$$ $$=-\int(1-\cos^2x)d(\cos{x})+6\int\sin^2xd(\sin{x})-12\int\cos^2xd(\cos{x})+$$ $$+8\int(1-\sin^2x)d(\sin{x})=$$ $$=-\cos{x}+\frac{1}{3}\cos^3x+2\sin^3x-4\cos^3x+8\sin{x}-\frac{8}{3}\sin^3x+C=$$ $$=8\sin{x}-\cos{x}-\frac{2}{3}\sin^3x-\frac{11}{3}\cos^3x+C.$$

Answer 5

Here is a very bizarre way to do this. I'm not claiming it's easier, but I just wanted to mention it because it's a little counterintuitive.

Let $\theta$ be an angle such that $$\cos \theta = \frac{1}{\sqrt{5}}, \quad \sin \theta = \frac{2}{\sqrt{5}}.$$ Then $$\sin x + 2 \cos x = \sqrt{5} \left(\cos \theta \sin x + \sin \theta \cos x \right) = \sqrt{5} \sin(x + \theta),$$ and $$\int (\sin x + 2 \cos x)^3 \, dx = 5 \sqrt{5} \int \sin^3 (x + \theta) \, dx = 5 \sqrt{5} \int \left(1 - \cos^2 (x + \theta)\right) \sin (x + \theta) \, dx.$$ Now letting $u = \cos (x + \theta)$, $du = -\sin(x + \theta) \, dx$, we obtain $$5 \sqrt{5} \int u^2 - 1\, du = 5 \sqrt{5} \left( \frac{u^3}{3} - u \right) = 5 \sqrt{5} \left( \frac{\cos^3 (x + \theta)}{3} - \cos (x + \theta)\right) + C.$$ We can substitute $\theta$ back into the expression; e.g., $$\frac{1}{3} (\cos x - 2 \sin x)^3 - 5 (\cos x - 2 \sin x) + C.$$