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Question Number 79059 by jagoll last updated on 22/Jan/20

$$$$$$$$$$\int \cos {}^2\left(\mathrm{x}\right)\sin {}^4\left(\mathrm{x}\right)\mathrm{dx}?$$

Commented by john santu last updated on 22/Jan/20

Commented by jagoll last updated on 22/Jan/20

$$\mathrm{thanks}\mathrm{sir}$$

Commented by mathmax by abdo last updated on 22/Jan/20

$$I=\int {cos}^{2}x{sin}^{2}x{sin}^{2}xdx$$$$=\int {\left(\frac{1}{2}sin\left(2x\right)\right)}^2\left(\frac{1-cos\left(2x\right)}{2}\right)dx$$$$=\frac{1}{8}\int \left(\frac{1-cos\left(4x\right)}{2}\right)(1-cos\left(2x\right))dx$$$$=\frac{1}{16}\int (cos\left(4x\right)-1)(cos\left(2x\right)-1)dx\Rightarrow$$$$16I=\int (cos\left(4x\right)cos\left(2x\right)-cos\left(4x\right)-cos\left(2x\right)+1)dx$$$$=\int \frac{1}{2}(cos\left(6x\right)+cos\left(2x\right))dx-\int cos\left(4x\right)dx-\int cos\left(2x\right)dx+x$$$$=\frac{1}{12}sin\left(2x\right)+\frac{1}{4}sin\left(2x\right)-\frac{1}{4}sin\left(4x\right)-\frac{1}{2}sin\left(2x\right)+x+c\Rightarrow$$$$I=\frac{1}{12\times 16}sin\left(2x\right)+\frac{1}{64}sin\left(2x\right)-\frac{1}{64}sin\left(4x\right)-\frac{1}{32}sin\left(2x\right)+\frac{x}{16}+C$$

Commented by mathmax by abdo last updated on 22/Jan/20

$$erroroftypo$$$$I=\frac{1}{12\times 16}sin\left(6x\right)+\frac{1}{64}sin\left(2x\right)-\frac{1}{64}sin\left(4x\right)-\frac{1}{32}sin\left(2x\right)+\frac{x}{16}+C$$

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