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education/math/MATH1220 (calc II)/Integration with Trig Identities.md
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@ -13,5 +13,11 @@ The below integration makes use of the following trig identities:
$$ \int\sin^4(x)\sin(x)dx$$ $$ \int\sin^4(x)\sin(x)dx$$
2. Rewrite $sin^4(x)$ to be $(\sin^2(x))^2$ to take advantage of the trig identity $1 - \cos^2(x) = \sin^2(x)$ 2. Rewrite $sin^4(x)$ to be $(\sin^2(x))^2$ to take advantage of the trig identity $1 - \cos^2(x) = \sin^2(x)$
$$ \int(\sin^2x)^2 \sin(x)dx$$ $$ \int(\sin^2x)^2 \sin(x)dx$$
3. Apply the above trig identity: 3. Apply the above trig identity and substitute $u$:
$$ \int(1) $$ \int(1 - u^2)^2 (-du) $$
4. Foil out and move negative out of integral:
$$ -\int(1 - 2u^2 + u^4)du $$
5. Take advantage of the distributive property of integrals:
$$ - (u - \frac{2}{3}u^3 + \frac{1}{5}u^5) + C $$
6. Substituting $\cos(x)$ back in for $u$, we get the evaluated (but not entirely simplified) integral:
$$-(\cos(x)- \frac{2}{3}\cos^3x + \frac{1}{5}\cos^5x) $$