Question

integration of 0 to pi cos square x dx ​

Answer 1

Solution :

Now, $\displaystyle \mathsf{\int_{0}^{\pi} cos^{2}x\:dx}$

= $\displaystyle \mathsf{\frac{1}{2}\int_{0}^{\pi} 2cos^{2}x\:dx}$

= $\displaystyle \mathsf{\frac{1}{2}\int_{0}^{\pi} (cos2x+1)dx}$

= $\displaystyle \mathsf{\frac{1}{2}\int_{0}^{\pi}cos2x\:dx+\frac{1}{2}\int_{0}^{\pi}dx}$

= $\displaystyle \mathsf{\frac{1}{2}[\frac{sin2x}{2}]_{0}^{\pi}+\frac{1}{2}[x]_{0}^{\pi}}$

= $\displaystyle \mathsf{\frac{1}{2}[\frac{sin2\pi}{2}-\frac{sin0}{2}]+\frac{1}{2}[\pi-0]}$

= $\displaystyle \mathsf{\frac{1}{2}(0-0)+\frac{1}{2}\pi}$

= $\displaystyle \mathsf{\frac{\pi}{2}}$ (Ans.)