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

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

\(\begin{array}{l} =60\cdot2\displaystyle\int\limits_0^{\frac{\pi}{2}}\left(3-4\sin^2x\right)\left(4\cos^2x-3\right)\cos x\,dx \end{array}\)

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

Let \(\sin x=t\) ⇒ \(\cos x\,dx=dt\)

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

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

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

\(\begin{array}{l} =120\left(3-\frac{16}{3}+\frac{16}{5}\right) \end{array}\)

\(\begin{array}{l} =120\left(\frac{45-80+48}{15}\right)=120\cdot\frac{13}{15}=104 \end{array}\)

The value of the integral is 104.