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Q16E

Expert-verifiedFound in: Page 326

Book edition
2nd

Author(s)
James Stewart

Pages
830 pages

ISBN
9781133112280

**Evaluate 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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