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Q16E
Expert-verifiedEvaluate the Integral: \(\int {{{\cos }^2}} x\sin 2xdx\)
To evaluate the integral \(\int {{{\cos }^2}} x\sin 2xdx\), we will use the identity \({\cos ^2}x = \frac{{1 + \cos 2x}}{2}\) followed by integration formulas
\(\int {{{\cos }^2}} x\sin 2xdx = \frac{{ - 1}}{4}\cos 2x - \frac{1}{{16}}\cos 4x + c\)
Let I = \(\int {{{\cos }^2}} x\sin 2xdx\)
As \({\cos ^2}x = \frac{{1 + \cos 2x}}{2}\)
\(I = \int {(\frac{{1 + \cos 2x}}{2}} )\sin 2xdx\)
\(\begin{aligned}{l}I &= \frac{1}{2}\int {(1 + \cos 2x)\sin 2xdx} \\ &= \frac{1}{2}\int {(\sin 2x + \sin 2x\cos 2x)dx} \end{aligned}\)
Now split the integral:
\(I = \frac{1}{2}\int {\sin 2xdx + \frac{1}{2}} \int {\sin 2x\cos 2xdx} \)
(1 Term) (2 Term)
\(\begin{aligned}{l}I &= \frac{1}{2}\int {\sin 2xdx + \frac{1}{4}} \int {2\sin 2x\cos 2xdx} \\ &= \frac{1}{2}\int {\sin 2xdx + \frac{1}{4}} \int {\sin 4xdx} \\ &= \frac{1}{2}(\frac{{ - 2\cos x}}{2}) + \frac{1}{4}(\frac{{ - \cos 4x}}{4}) + c\\\end{aligned}\)
\(I = \frac{{ - \cos 2x}}{4} - \frac{1}{{16}}\cos 4x + c\)
Hence value of integral:
\(\int {{{\cos }^2}} x\sin 2xdx = \frac{{ - 1}}{4}\cos 2x - \frac{1}{{16}}\cos 4x + c\)
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