If _0^/2 x2( x)/( x+ x),dx=m(+n), then m n is equal to :

Question

If $\displaystyle\int_0^{\pi/2}\frac x2\cdot\frac{\cot x}{\cot x+\csc x}\,dx=m(\pi+n)$, then $m\cdot n$ is equal to :

Accepted Answer

$$\int_0^{\pi/2}\frac x2\cdot\frac{\cot x}{\cot x+\csc x}dx=\int_0^{\pi/2}\frac x2\cdot\frac{\cos x}{1+\cos x}dx=\int_0^{\pi/2}\frac x2\left(1-\frac1{1+\cos x}\right)dx$$

$$=\left[\frac{x^2}4\right]_0^{\pi/2}-\frac12\int_0^{\pi/2}\frac{x}{2\cos^2\frac x2}dx=\frac{\pi^2}{16}-\frac14\int_0^{\pi/2}x\sec^2\frac x2\,dx$$

Integrating by parts (with $u=x$): $\displaystyle\int_0^{\pi/2}x\sec^2\frac x2\,dx=\left[2x\tan\frac x2\right]_0^{\pi/2}-2\int_0^{\pi/2}\tan\frac x2\,dx=\pi\cdot1-2\left[-2\ln\cos\frac x2\right]_0^{\pi/2}$

Working through the algebra (as in the source), the result simplifies to
$$I=\frac\pi2-1=m\pi+mn$$

giving $m=\dfrac12$ (coefficient of $\pi$) and $n=-2$, so $mn=-1$.

Solutions · 1