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Question Number 187251 by horsebrand11 last updated on 15/Feb/23

$$\int \frac{\cos 9x}{\cos 4x.\cos 2x}dx=?$$

Commented by MJS_new last updated on 15/Feb/23

$$=4\int (1-{4\sin }^{2}x)\cos xdx-$$$$-\int \frac{\cos x}{1-{2\sin }^{2}x}dx-$$$$-\frac{1}{4}\int \frac{(1-{2\sin }^{2}x)\cos x}{{\sin }^{4}x-{\sin }^{2}x+1/8}dx$$$$\mathrm{and}\mathrm{these}\mathrm{are}\mathrm{easy}\mathrm{to}\mathrm{solve}$$

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