Question

$$\int (1+x^2)\cos 2xdx$$

Answer 1

$$\text{ Let I}=\int (1+x^2)\cdot \cos 2x\cdot dx$$
$$=\int \text{ cos 2x dx }+\int x^2\cdot \text{ cos 2x dx }$$
$$=\frac{\sin 2x}{2}+I_1{\text{ where I}}_1=\int x^2\text{ cos 2x dx }...\left(1\right)$$
$$I_1=\int {x^2}_I{\cos }_{II}2x\text{ dx }$$
$$=x^2\int \text{ cos 2x dx }-\int \{\frac{d}{dx}\left(x^2\right)\int \text{ cos 2x dx }\}dx$$
$$=x^2\cdot \frac{\sin 2x}{2}-\int \frac{2x\times \sin 2x}{2}dx$$
$$=\frac{x^2\cdot \sin 2x}{2}-\int x_I\cdot {\sin }_{II}2x\text{ dx }$$
$$=\frac{x^2\cdot \sin 2x}{2}-[x\int \text{ sin 2x dx }-\int \{\frac{d}{dx}\left(x\right)\int \text{ sin 2x dx}\}dx]$$
$$=\frac{x^2\cdot \sin 2x}{2}-[x\left(\frac{-\cos 2x}{2}\right)-\int 1\cdot \left(\frac{-\cos 2x}{2}\right)dx]$$
$$=\frac{x^2\cdot \sin 2x}{2}+\frac{x\cdot \cos 2x}{2}-\frac{\sin 2x}{4}...\left(2\right)$$
$$\text{ From }\left(1\right)\text{ and }\left(2\right)$$
$$\therefore I=\frac{\sin 2x}{2}+\frac{x^2\sin 2x}{2}+\frac{x\cos 2x}{2}-\frac{\sin 2x}{4}+C$$
$$=(x^2+1)\frac{\sin 2x}{2}+\frac{x\cos 2x}{2}-\frac{\sin 2x}{4}+C$$