Question

Evaluate the definite integrals.
${\int }_0^{\frac{\pi }{4}}sin2xdx.$

Answer 1

Let $I={\int }_0^{\frac{\pi }{4}}sin2xdx={\int }_0^{\frac{\pi }{4}}2sinxcosxdx$
$[\because$ $\text{sin 2x =2 sin x cos x}$]
Put $sinx=t\Rightarrow cosx=\frac{dt}{dx}\Rightarrow dx=\frac{dt}{cosx}$
Upper limit, at $x=\frac{\pi }{4}\Rightarrow t=sin\frac{\pi }{4}=\frac{1}{\sqrt{2}}$
Lower limit, at $x=0\Rightarrow t$ =$sin0=0$

$\therefore I=2{\int }_0^{\frac{1}{\sqrt{2}}}tcosx\frac{dt}{cosx}=2{\int }_0^{\frac{1}{\sqrt{2}}}tdt=2{\left[\frac{t^2}{2}\right]}_0^{\frac{1}{\sqrt{2}}}={\left(\frac{1}{\sqrt{2}}\right)}^2=\frac{1}{2}$

Note If we change the integral function by considerable another variable, then remember to change the limit or after integration change into the given variable and then apply the limit.