Question

The value of the integral $\begin{array}{l} \displaystyle\int\limits_0^{\frac{\pi}{2}}60\frac{\sin\left(6x\right)}{\sin x}dx \end{array}$ is equal to ______.

Answer 1

$\begin{array}{l} I=\displaystyle\int\limits_0^{\frac{\pi}{2}}60\cdot\frac{\sin6x}{\sin x}dx \end{array}$

$\begin{array}{l} =60.2\displaystyle\int\limits_0^{\frac{\pi}{2}}\left(3-4\sin^2x\right)\left(4\cos^2x-3\right)\cos xdx\end{array}$

$\begin{array}{l} =120\displaystyle\int\limits_0^{\frac{\pi}{2}}\left(3-4\sin^2x\right)\left(1-4\sin^2x\right)\cos xdx\end{array}$

Let sin x = t ⇒ cos x dx = dt

$\begin{array}{l} =120\displaystyle\int\limits_0^1\left(3-4t^2\right)\left(1-4t^2\right)dt\end{array}$

$\begin{array}{l} =120\displaystyle\int\limits_0^1\left(3-16t^2+16t^4\right)dt\end{array}$

\begin{array}{l} =120\left[3t-\frac{16t^3}{3}+\frac{16t^5}{5}\right]_0^1\\\= 104 \end{array}