Final answer to the problem
$\frac{1}{16}\sin\left(2x\right)+\frac{1}{8}x+\frac{-\cos\left(x\right)^{3}\sin\left(x\right)}{4}+C_0$
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Step-by-step Solution
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Intermediate steps
1
Rewrite the trigonometric expression $\cos\left(x\right)^2\sin\left(x\right)^2$ inside the integral
$\int\left(\cos\left(x\right)^2-\cos\left(x\right)^{4}\right)dx$
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2
Expand the integral $\int\left(\cos\left(x\right)^2-\cos\left(x\right)^{4}\right)dx$ into $2$ integrals using the sum rule for integrals, to then solve each integral separately
$\int\cos\left(x\right)^2dx+\int-\cos\left(x\right)^{4}dx$
Intermediate steps
3
The integral $\int\cos\left(x\right)^2dx$ results in: $\frac{1}{2}x+\frac{1}{4}\sin\left(2x\right)$
$\frac{1}{2}x+\frac{1}{4}\sin\left(2x\right)$
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4
Gather the results of all integrals
$\frac{1}{4}\sin\left(2x\right)+\frac{1}{2}x+\int-\cos\left(x\right)^{4}dx$
Intermediate steps
5
The integral $\int-\cos\left(x\right)^{4}dx$ results in: $\frac{-\cos\left(x\right)^{3}\sin\left(x\right)}{4}-\frac{3}{4}\left(\frac{1}{2}x+\frac{1}{4}\sin\left(2x\right)\right)$
$\frac{-\cos\left(x\right)^{3}\sin\left(x\right)}{4}-\frac{3}{4}\left(\frac{1}{2}x+\frac{1}{4}\sin\left(2x\right)\right)$
Explain this step further
6
Gather the results of all integrals
$\frac{1}{4}\sin\left(2x\right)+\frac{1}{2}x-\frac{3}{4}\left(\frac{1}{2}x+\frac{1}{4}\sin\left(2x\right)\right)+\frac{-\cos\left(x\right)^{3}\sin\left(x\right)}{4}$
7
As the integral that we are solving is an indefinite integral, when we finish integrating we must add the constant of integration $C$
$\frac{1}{4}\sin\left(2x\right)+\frac{1}{2}x-\frac{3}{4}\left(\frac{1}{2}x+\frac{1}{4}\sin\left(2x\right)\right)+\frac{-\cos\left(x\right)^{3}\sin\left(x\right)}{4}+C_0$
Intermediate steps
$\frac{1}{16}\sin\left(2x\right)+\frac{1}{8}x+\frac{-\cos\left(x\right)^{3}\sin\left(x\right)}{4}+C_0$
Explain this step further
Final answer to the problem
$\frac{1}{16}\sin\left(2x\right)+\frac{1}{8}x+\frac{-\cos\left(x\right)^{3}\sin\left(x\right)}{4}+C_0$