Integrals Involving Trig Functions — Question 9

Question 9 – Solution

Begin with the power-reduction identity $$\cos^2(x)=\frac{1+\cos(2x)}2.$$ Squaring both sides, $$\begin{align*} \cos^4(x) &=\left(\frac{1+\cos(2x)}2\right)^2\\ &=\frac14\left(1+2\cos(2x)+\cos^2(2x)\right). \end{align*}$$ Apply power reduction again: $$\cos^2(2x)=\frac{1+\cos(4x)}2.$$ Therefore, $$\begin{align*} \cos^4(x) &=\frac14\left(1+2\cos(2x)+\frac{1+\cos(4x)}2\right)\\ &=\frac38+\frac12\cos(2x)+\frac18\cos(4x). \end{align*}$$ Matching this with the required form gives $$\boxed{A=\frac38,\qquad B=\frac12}.$$ Now integrate the identity term by term: $$\begin{align*} \int\cos^4(x)\,dx &=\int\left(\frac38+\frac12\cos(2x)+\frac18\cos(4x)\right)\,dx\\ &=\frac{3x}{8}+\frac14\sin(2x)+\frac1{32}\sin(4x)+C. \end{align*}$$ Hence, $$\boxed{\displaystyle \int\cos^4(x)\,dx =\frac{3x}{8}+\frac14\sin(2x)+\frac1{32}\sin(4x)+C}.$$