Question

Integration of \(\int\limits_0^\pi {co{s^4}(2t)dt} \)

Short answer

\(\frac{{3\pi }}{8}\)is the answer to the given Question

Step-by-step solution

Concept Used

\(\)\({\cos ^4}(2t) = {\left( {{{\cos }^2}(2t)} \right)^2}\)

Calculating Integral

\(\int\limits_0^\pi {{{\left( {{{\cos }^2}(2t)} \right)}^2}\,dt} \) \(\left( \begin{aligned}{l}{\cos ^2}(\theta ) = \frac{{1 + \cos \,2\theta }}{2}\\{\cos ^2}(t) = \frac{{1 + \cos \,2t}}{2}\end{aligned} \right)\)

\(\begin{aligned}{l} &= {\int\limits_0^\pi {\left( {\frac{{1 + \cos 4t}}{2}} \right)} ^2}dt\\ &= \frac{1}{4}\int\limits_0^\pi {\left( {1 + {{\cos }^2}4t + 2\cos 4t} \right)} dt\end{aligned}\)\(\) \({(a + b)^2} = ({a^2} + {b^2} + 2ab)\)

\(\begin{aligned}{l} = \frac{1}{4}{}_0^\pi \left( {\left( {t + 2\frac{{\sin 4t}}{4}} \right) + \int\limits_0^\pi {\frac{{1 + \cos 8t}}{2}} } \right) \times \frac{1}{4}\\ = \frac{1}{4}\left( {t + \frac{{\sin 4t}}{2} + \frac{t}{2} + \frac{{\sin 8t}}{{16}}} \right)_0^\pi \\\left( {\therefore \int {dt = {t_1}\int {\cos 2t = \frac{{\sin 2t}}{2}} } } \right)\end{aligned}\)

 Now put the Upper limit-Lower limit

\( = \frac{1}{4}\left( {\frac{{3t}}{2}} \right)_0^\pi \) \(\left( {\sin \,4\pi = \sin \,8\pi = \sin \,\pi = \sin 0 = 0} \right)\)

\(\begin{aligned}{l} &= \frac{1}{4}\left( {\frac{{3\pi }}{2}} \right)\\ &= \left( {\frac{{3\pi }}{8}} \right)\end{aligned}\)

\(\left( {\frac{{3\pi }}{8}} \right)\) is the final answer to the question.