A calculus problem by Hamza A

Substitute $x=a\sin \theta$ such that $\mathrm{d}x=\cos \theta~ \mathrm{d}\theta$ and then using $1-\sin^2\theta=\cos^2\theta$ .

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$\mathfrak{I}=\displaystyle\int_0^{\frac{\pi}{2}}\dfrac{\cos \theta~\mathrm{d}\theta}{\sin \theta +cos \theta }$

Using $\displaystyle\int_a^b f(x)\mathrm{d}x=\displaystyle\int_a^b f(a+b-x)\mathrm{d}x$ and adding to get:

$2\mathfrak{I}=\displaystyle\int_0^{\frac{\pi}{2}}\dfrac{\cancel{\cos \theta+\sin \theta}~\mathrm{d}\theta}{\cancel{\sin \theta +cos \theta} }$

$\implies 2\mathfrak I=\displaystyle\int_0^{\frac{\pi}{2}}\mathrm{d}\theta$

$\Large \implies \mathfrak I=\dfrac{\pi}{4}$

$\huge \therefore\boxed{ B=4}$