$$\int\left(cos\left(9t-x\right)\right)dx$$

$$\left(4x-3\right)-\left(8x+9\right)$$

$$-y^2\left(-y^2\right)$$

$$\frac{\left(cos^2x\:-\:2cosx\:+\:1\right)}{sin^2x}=\frac{\left(1-cosx\right)}{1+cosx}$$

$$24^2\cdot34$$

$$cosa+1=4cosa+1$$