Question

The value of ${\int }_0^{\frac{\mathrm{\pi }}{2}}\frac{\left[\left(\cos \right(3x)+1)\right]}{(2\cos x-1)}dx=$

• A. $2$
• B. $1$
• C. $\frac{1}{2}$
• D. $0$

Answer 1

The correct option is B

$1$

Explanation for correct option

Given: ${\int }_0^{\frac{\mathrm{\pi }}{2}}\frac{\left[\left(\cos \right(3x)+1)\right]}{(2\cos x-1)}dx$

Let ${\int }_0^{\frac{\mathrm{\pi }}{2}}\frac{\left[\left(\cos \right(3x)+1)\right]}{(2\cos x-1)}dx=I$

$\begin{array}{l}\begin{array}{l}={\int }_0^{\frac{\pi }{2}}\left(\frac{1+\cos \left(3x\right)}{2\cos \left(x\right)-1}\right)dx\\ ={\int }_0^{\frac{\pi }{2}}\frac{\cos \left(3x\right)-\cos \left(\frac{3\pi }{3}\right)}{2\left(\cos \left(x\right)-\frac{1}{2}\right)}dx\\ ={\int }_0^{\frac{\pi }{2}}\frac{\cos \left(3x\right)-\cos \left(\frac{3\pi }{3}\right)}{2\left(\cos \right(x)-\cos \left(\frac{\mathrm{\pi }}{3}\right)}dx\left[as\cos \left(\frac{\mathrm{\pi }}{3}\right)=\frac{1}{2}\right]\\ ={\int }_0^{\frac{\pi }{2}}\frac{\left(4{\cos }^3\right(x)-3\cos (x\left)\right)-(4{\cos }^3\left(\frac{\pi }{3}\right)-3\cos \left(\frac{\pi }{3}\right))}{2\left(\cos \right(x)-\cos \left(\frac{\pi }{3}\right))}dx\left[as\cos \left(3x\right)=4{\cos }^3\left(x\right)-3\cos \left(x\right)\right]\\ ={\int }_0^{\frac{\pi }{2}}\frac{4\left({\cos }^3\left(x\right)-{\cos }^3\left(\frac{\pi }{3}\right)\right)}{2\left(\cos \left(x\right)-\cos \left(\frac{\pi }{3}\right)\right)}-\frac{3\left(\cos \left(x\right)-\cos \left(\frac{\pi }{3}\right)\right)}{2\left(\cos \left(x\right)-\cos \left(\frac{\pi }{3}\right)\right)}dx\\ ={\int }_0^{\frac{\pi }{2}}\frac{2\left(\cos \left(x\right)-\cos \left(\frac{\pi }{3}\right)\right)\left({\cos }^2\left(x\right)+{\cos }^2\left({\displaystyle \frac{\mathrm{\pi }}{3}}\right)+\cos \left(x\right)·\cos \left({\displaystyle \frac{\mathrm{\pi }}{3}}\right)\right)}{\left(\cos \left(x\right)-\cos \left(\frac{\pi }{3}\right)\right)}-\frac{3\left(\cos \left(x\right)-\cos \left(\frac{\pi }{3}\right)\right)}{2\left(\cos \left(x\right)-\cos \left(\frac{\pi }{3}\right)\right)}dx\left[as(a^3-b^3)=(a-b)(a^2+b^2+ab)\right]\\ ={\int }_0^{\frac{\pi }{2}}2\left({\cos }^2\left(x\right)+\frac{1}{2}+\cos x\right)dx-\frac{3}{2}{\int }_0^{\frac{\pi }{2}}[x].dx\\ ={\int }_0^{\frac{\pi }{2}}\left(1+\cos \left(2x\right)+\frac{1}{2}+\cos x\right)dx-\frac{3}{2}\left(\frac{\pi }{2}-0\right)\left[\cos \left(2x\right)=2{\cos }^2\left(x\right)-1\right]\\ ={\int }_0^{\frac{\pi }{2}}\left(\cos \left(2x\right)+\frac{3}{2}+\cos \left(x\right)\right)dx-\frac{3\pi }{4}\\ ={\left[\frac{\sin \left(2x\right)}{4}+\frac{3}{2}x+\sin \left(x\right)\right]}_0^{\frac{\pi }{2}}-\frac{3\pi }{4}\\ =\frac{\sin 2\left(\frac{\pi }{2}\right)}{2}-\frac{\sin 0}{2}+\frac{3}{2}\left(\frac{\pi }{2}-0\right)+\left(\sin \frac{\pi }{2}-\sin 0\right)-\frac{3\pi }{4}\\ =\frac{\sin \left(\mathrm{\pi }\right)}{2}+\frac{3\mathrm{\pi }}{4}+1-\frac{3\mathrm{\pi }}{4}\\ =0+1-0\\ =1\end{array}\end{array}$Hence, option B is correct.