Mathematics

Find the integrals of the functions.
i) $$sin^3 \, x \, cos^3 \, x$$
ii) $$sin \, x \, sin \, 2x \, sin \, 3x$$
iii) $$sin \, 4x \, sin \, 8x$$
iv) $$\dfrac{1 - cos \, x}{1 + cos \, x}$$
v) $$\dfrac{cos \, x}{1 + cos \, x}$$


SOLUTION
(i) $$I=\displaystyle\int \sin^3x\cos^3xdx=\dfrac{1}{8}\displaystyle\int(\sin 2x)^3dx\Rightarrow \dfrac{1}{32}\displaystyle\int (3\sin 3x-4\sin(6x))dx$$
$$(\sin 3x=3\sin x-4\sin^3x)$$
$$I=\dfrac{1}{32}\displaystyle\int 3\sin 2xdx-\dfrac{1}{32}4\sin (6x)dx$$
$$=\dfrac{3}{32.2}\times -(\cos 2x)+\dfrac{4}{6.32}\cos (6x)+C$$
$$I=\dfrac{\cos (6x)}{48}-\dfrac{3\cos (2x)}{64}+C$$.
(ii) $$\displaystyle\int \sin x\sin 2x\sin 3xdx$$
$$\Rightarrow \displaystyle\int\dfrac{1}{2}\left[\cos(3x-x)-\cos (3x+x)\right]\sin 2xdx$$
$$\Rightarrow \displaystyle\int \dfrac{1}{2}\sin 2x\cos 2xdx-\displaystyle\dfrac{1}{2}\sin 2x\cos 4xdx$$
$$\Rightarrow \dfrac{1}{4}\displaystyle\int \sin 4xdx-\dfrac{1}{4}\displaystyle\int (\sin (2x+4x)+\sin (2x-4x)dx)$$
$$\Rightarrow -\dfrac{1}{16}\cos 4x-\dfrac{1}{4}\displaystyle\int \sin 6xdx+\dfrac{1}{4}\displaystyle\int \sin 2xdx$$
$$I\Rightarrow \dfrac{1}{24}\cos 6x-\dfrac{1}{16}\cos 4x-\dfrac{1}{8}\cos 2x+C$$
(iii) $$\displaystyle\int \sin 4x\sin 8xdx$$
$$\Rightarrow \displaystyle\int \dfrac{1}{2}\left[\cos (8x-4x)-\cos (8x+4x)\right]dx$$
$$\Rightarrow \displaystyle\int \dfrac{1}{2}(\cos 4x-\cos 12x)dx$$
$$\Rightarrow \dfrac{1}{8}\sin 4x-\dfrac{1}{24}\sin 12x+C$$.
(iv) $$\displaystyle\int \dfrac{(1-\cos x)}{(1+\cos x)}dx$$
$$\Rightarrow \displaystyle\int \dfrac{2\sin^2x/_{2}}{2\cos^2 x/_{2}}dx$$
$$\Rightarrow \displaystyle\int \dfrac{1-\cos^2x/_{2}}{\cos^2x/_{2}}dx$$
(iv) $$\displaystyle\int \dfrac{(1-\cos x)}{(1+\cos x)}dx$$
$$\Rightarrow \displaystyle\int \dfrac{2\sin^2x/_{2}][2\cos^2x/_{2}}dx$$
$$\Rightarrow \displaystyle\int \dfrac{1-\cos^2x/_{2}}{\cos^2x/_{2}}dx$$
$$\Rightarrow \displaystyle\int \sin^2x/_{2}dx=\displaystyle\int dx$$
$$\Rightarrow 2\tan x/_{2}-x+C$$
(vi) $$\displaystyle\int \dfrac{\cos x}{1+\cos x}dx$$
$$\Rightarrow \displaystyle\int \left(\dfrac{1+\cos x}{cos x+1}-\dfrac{1}{\cos x+1}\right)dx$$
$$\Rightarrow \displaystyle\int dx-\displaystyle\int \dfrac{1}{1+\cos x}dx$$
$$\Rightarrow x-\dfrac{1}{2}\displaystyle\int \sec^2x/_{2}dx$$
$$\Rightarrow x-\dfrac{1}{2}\times 2\tan x/_{2}+C$$
$$\Rightarrow x-\tan x/_{2}+C$$.
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