Mathematics

Integrate:
$$\int \dfrac{\cos 4x}{\sin^2x}dx$$


SOLUTION
$$I=\displaystyle\int\dfrac{\cos 4x}{\sin^{2}x}dx$$

$$=\displaystyle\int \dfrac{\cos^{2}2x-\sin^{2}2x}{\sin^{2}x}dx$$

$$=\displaystyle\int\dfrac{(\cos^{2}x-\sin^{2}x)^2-4\sin^{2}x\cos^{2}x}{\sin^{2}x}dx$$

$$=\displaystyle\int\dfrac{\cos^{4}x+\sin^{4}x-6\sin^{2}x\cos^{2}x}{\sin^{2}x}dx$$

$$=\displaystyle\int\dfrac{\sin^{4}x+\cos^{2}x(1-7\sin^{2}x)dx}{\sin^{2}x}$$

$$[\because \sin^{2}x+\cos^{2}x=1]$$

$$\displaystyle\int \sin^{2}xdx+\int\dfrac{\cos^{2}x(1-7\sin^{2}x)dx}{\sin^{2}x}$$

$$=I_{1}+I_{2}$$ ............. $$(1)$$

$$I_{1}=\displaystyle\int\sin^{2}x dx=\int \dfrac{(1-\cos 2x)}{2}dx$$

$$=\dfrac{x}{2}-\dfrac{\sin 2x}{4}$$ ....... $$(2)$$

$$I_{2}=\displaystyle\int\dfrac{\cos^{2}x(1-7\sin^{2}x)}{\sin^{2}x}dx$$

$$=\displaystyle\int\cot^{2}dxd-7\int \cos^{2}x dx$$

$$=\displaystyle\int (\csc^{2}x-1)dx-\dfrac{7}{2}\int (1+\cos 2x)dx$$

$$=-\cot x-x\dfrac{-7}{2}x-\dfrac{7}{4}\sin 2x$$

$$=-\cot x-\dfrac{7}{4}\sin 2x-\dfrac{9}{2}x$$ ........ $$(3)$$

$$I=I_{1}+I_{2}$$

$$\Rightarrow I=-\cot x-2\sin 2x-4x+C$$
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