Question

Evaluate: ${\int }_0^{\pi }\frac{4xsinx}{1+co{s}^{2}x}dx$.

OR

Evaluate: $\int \frac{x+2}{\sqrt{x^2+5x+6}}dx$.

Answer 1

Let $I={\int }_0^{\pi }\frac{4xsinx}{1+co{s}^{2}x}dx\dots \left(i\right)$

Using ${\int }_0^{a}f\left(x\right)dx={\int }_0^{a}f(a-x)dx,I={\int }_0^{\pi }\frac{sinx}{1+co{s}^2(\pi -x)}dx={\int }_0^{\pi }\frac{4(\pi -x)sinx}{1+co{s}^{2}x}dx\dots \left(ii\right)$

Adding (i) and (ii), we get $21=4\pi {\int }_0^{\pi }\frac{sinx}{1+co{s}^2(\pi -x)}dx\Rightarrow 1=2\pi {\int }_0^{\pi }\frac{sinx}{1+co{s}^{2}x}dx$

Put $cosx=t\Rightarrow sinxdx=-dt$. Also when, $x=0\Rightarrow t=1andwhenx=\pi \Rightarrow t=-1$

$\therefore I=2\pi {\int }_1^{-1}\frac{-dt}{1+t^2}\Rightarrow 1=2\pi [ta{b}^{-1}t{]}_{-1}^1=2\pi [ta{n}^{-1}\left(1\right)-ta{n}^{-1}(-1)]$

$\Rightarrow I=2\pi [\frac{\pi }{4}-(-\frac{\pi }{4})]\therefore I={\pi }^2$

OR

Let $I=\int \frac{x+2}{\sqrt{x^2+5x+6}}dx\Rightarrow I=\int \frac{\frac{1}{2}(2x+5)-\frac{1}{2}}{\sqrt{x^2+5x+6}}dx$

$\Rightarrow I=\frac{1}{2}\int \frac{(2x+5)}{\sqrt{{}^2+5x+6}}dx-\frac{1}{2}\int \frac{1}{\sqrt{x^2+5x+6}}dx$

$\Rightarrow I=\frac{1}{2}\left[2\sqrt{x^2+5x+6}\right]-\frac{1}{2}\int \frac{1}{\sqrt{{(x+\frac{5}{2})}^2-{\left(\frac{1}{2}\right)}^2}}dx$

$\Rightarrow I=\sqrt{x^2+5x+6}-\frac{1}{2}log|(x+\frac{5}{2})+\sqrt{{(x+\frac{5}{2})}^2-{\left(\frac{1}{2}\right)}^2}|+C$

or, $I=\sqrt{x^2+5x+6}-\frac{1}{2}log|x+\frac{5}{2}+\sqrt{x^2+5x+6}|+C$