Mathematics

$$I = \int
{{{\cos }^4}x.dx} $$


SOLUTION
$$\displaystyle \int \cos^4 x dx$$
Now $$\cos^4 x = \cos^3 x . \cos x$$
$$\displaystyle\int \cos^4 x dx = \int \cos^3 x d (\sin x)$$
$$\displaystyle\int  \cos^4 x dx = \sin  x \cos^3 x - \int \sin x d(\cos^3 x) \rightarrow $$ By parts
$$\displaystyle\int \cos^4 x dx = \sin x \cos^3 x + 3 \int sin^2 x \cos^2 x \, dx$$
$$\displaystyle\int \cos^4 x dx = \sin x \cos^3 x + 3 \int (1 - \cos^2 x) \cos^2 x \, dx$$
$$\displaystyle\int \cos^4 x dx = \sin x \cos^3 x + 3 \int \cos^2 x dx - 3 \int \cos^4 x \, dx$$
$$4 \displaystyle\int \cos^4 x dx = \sin x \cos^3 x + 3 \int \cos^2 x \, dx$$
$$\displaystyle\int \cos^4 x dx = \dfrac{\sin x \, \cos^3 x}{4} + \dfrac{3}{4} \int \cos^2 x dx$$
Now, $$\displaystyle\int \cos^2 x dx = \int \left(\dfrac{\cos 2x + 1}{2} \right) dx$$
$$\therefore \displaystyle\int \cos^2 x \, dx = \dfrac{\sin 2x }{4} + \dfrac{x}{2}$$
$$\therefore \displaystyle\int \cos^4 x \, dx = \dfrac{\sin x \cos^3 x}{4} + \dfrac{3}{4} \left(\dfrac{\sin 2x}{4} + \dfrac{x}{2} \right)$$
$$= \displaystyle\int \cos^4 x \, dx = \dfrac{\sin x \, \cos^3 x}{4} + \dfrac{3 \, \sin 2x}{16} + \dfrac{3x}{8}$$
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Subjective Medium Published on 17th 09, 2020
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