Question

$\mathrm{If}F\left(x\right)=\int \frac{\mathrm{dx}}{1+\mathrm{sinx}+\mathrm{cosx}},\mathrm{and}F\left(0\right)=0,\mathrm{then}\mathrm{the}\mathrm{value}\mathrm{of}F\left(\frac{\pi }{2}\right)\mathrm{is}$.

• A. ${\log }_{e}2$
• B. ${\log }_{e}3$
• C. 0.0

Answer 1

The correct option is A ${\log }_{e}2$

To solve integral of the form $\int \frac{\mathrm{dx}}{a+b\mathrm{cosx}+c\mathrm{sinx}}$, we write $\mathrm{sinx}\mathrm{and}\mathrm{cosx}\mathrm{in}\mathrm{terms}\mathrm{of}\tan \frac{x}{2},\mathrm{and}\mathrm{then}\mathrm{substitute}t=\tan \frac{x}{2}.$
Now, putting $\mathrm{sinx}=\frac{2\tan \frac{x}{2}}{1+{\tan }^2\frac{x}{2}}\mathrm{and}\mathrm{cosx}=\frac{1-{\tan }^2\frac{x}{2}}{1+{\tan }^2\frac{x}{2}}.$
We have:
$I=\int \frac{\mathrm{dx}}{1+\mathrm{sinx}+\mathrm{cosx}}=\int \frac{\mathrm{dx}}{1+\frac{2\tan \frac{x}{2}}{1+{\tan }^2\frac{x}{2}}+\frac{1-{\tan }^2\frac{x}{2}}{1+{\tan }^2\frac{x}{2}}}=\int \frac{\left(1+{\tan }^2\frac{x}{2}\right)\mathrm{dx}}{1+{\tan }^2\frac{x}{2}+2\tan \frac{x}{2}+1-{\tan }^2\frac{x}{2}}=\int \frac{{\sec }^2\frac{x}{2}\mathrm{dx}}{2+2\tan \frac{x}{2}}\mathrm{Now},\mathrm{susbtituting}t=\tan \frac{x}{2,}\mathrm{we}\mathrm{get}\mathrm{dt}=\frac{1}{2}{\sec }^2\frac{x}{2}\mathrm{dx}\mathrm{Thus},\mathrm{our}\mathrm{integral}\mathrm{becomes}:I=\int \frac{2\mathrm{dt}}{2(1+t)}\Rightarrow I=\int \frac{\mathrm{dt}}{(1+t)}=\ln \left(1+t\right|+C\mathrm{Substituting}\mathrm{back}t=\tan \frac{x}{2},\mathrm{we}\mathrm{get}I=\ln \left(1+\tan \frac{x}{2}\right|+C=F\left(x\right)\mathrm{Also},F\left(0\right)=0\Rightarrow \ln \left(1+\tan 0\right|+C=0\Rightarrow \ln \left(1\right|+C=0\Rightarrow C=0\mathrm{Thus},F\left(x\right)=\ln \left(1+\tan \frac{x}{2}\right|\mathrm{Also},F\left(\frac{\pi }{2}\right)=\ln \left(1+\tan \frac{\pi }{4}\right|=\ln \left(2\right|={\log }_{e}2$