Mathematics - Indefinite Integration Question with Solution | TestHub

Let $t=\sin x+\cos x$. Then $dt=(\cos x-\sin x)dx$. Squaring $t$, we get $t^2=(\sin x+\cos x{)}^2={\sin }^{2}x+{\cos }^{2}x+2\sin x\cos x=1+\sin (2x)$. From this, we have $\sin (2x)=t^2-1$. Substituting these into the integral:$$\int \frac{dt}{\sqrt{8-(t^2-1)}}=\int \frac{dt}{\sqrt{9-t^2}}$$ This is a standard integral of the form $\int \frac{dx}{\sqrt{a^2-x^2}}={\sin }^{-1}\left(\frac{x}{a}\right)+C$. Here $a=3$, so the integral is ${\sin }^{-1}\left(\frac{t}{3}\right)+C$. Substituting back $t=\sin x+\cos x$: $$={\sin }^{-1}\left(\frac{\sin x+\cos x}{3}\right)+C$$