Question

Solve this.

Solve this. $\int {\text{sin}}^2{\text{x cos}}^4\text{x dx}$

Answer 1

dear student

$$\begin{gathered} I=\int {\sin }^{2}x{\cos }^{4}xdx \\ =\int \frac{1-\cos 2x}{2}{\left(\frac{1+\cos 2x}{2}\right)}^{2}dx \\ =\frac{1}{8}\int (1-\cos 2x)(1+{\cos }^{2}2x+2\cos 2x)dx \\ =\frac{1}{8}\int (1+{\cos }^{2}2x+2\cos 2x-\cos 2x-{\cos }^{3}2x-2{\cos }^{2}2x)dx \\ =\frac{1}{8}\int (1-{\cos }^{2}2x+\cos 2x-{\cos }^{3}2x)dx \\ =\frac{1}{8}\int (1-\frac{1+\cos 4x}{2}+\cos 2x-{\cos }^{2}2x.\cos 2x)dx \\ =\frac{1}{8}(x/2-\sin 4x/8+\sin 2x/2-\int (1-{\sin }^{2}2x\left)\cos 2xdx\right)+c \\ =\frac{1}{8}(x/2-\sin 4x/8+\sin 2x/2-\frac{1}{2}(\sin 2x-\frac{{\sin }^{3}2x}{3}\left)\right)+c \end{gathered}$$

regards