Question

Evaluate the definite integrals.
${\int }_0^{\frac{\pi }{4}}\frac{sinx+cosx}{9+16sin2x}dx.$

Answer 1

$I={\int }_0^{\frac{\pi }{4}}\frac{sinx+Cosx}{9+16sin2x}dx.$
We know that $(sinx-cosx{)}^2=si{n}^{2}x+co{s}^{2}x-2sinxcosx$
$\Rightarrow (sinx-cosx{)}^2=1-sin2x\text{[}\because si{n}^{2}x\text{+}co{s}^{2}x\text{= 1 and 2 sin x cos x = sin 2x]}\Rightarrow sin2x=1-(sinx-cosx{)}^2\text{Put sin x - cos x}=t\Rightarrow cosx+sinx=\frac{dt}{dx}\Rightarrow dx=\frac{dt}{cosx+sinx}Whenx=0,t=0-cos0=-1,X=\frac{\pi }{4},t=sin\frac{\pi }{4}-cos\frac{\pi }{4}=0\therefore I={\int }_{-1}^0\frac{sinx+cosx}{9+16(1-t^2)}\frac{dt}{cosx+sinx}={\int }_{-1}^0\frac{1}{9+16(1-t^2)}dt=\frac{1}{16}.\frac{1}{2.\frac{5}{4}}{\left[log\left|\frac{\frac{5}{4}+t}{\frac{5}{4}-t}\right|\right]}_{-1}^0(\because \int \frac{1}{a^2-x^2}dx=\frac{1}{2a}log\left|\frac{a+x}{a-x}\right|)=\frac{1}{40}{\left[log\left|\frac{5+4t}{5-4t}\right|\right]}_{-1}^0=\frac{1}{40}(log1-log\frac{1}{9})=\frac{1}{40}[log1-(log1-log9\left)\right]=\frac{1}{40}log9[\because log\left(\frac{m}{n}\right)=logm-logn]$