a) Evaluate: dx5 + 4 x [3] b) Solve the differential equation 1 x^2\,dy + 1 y^2\,dx = 0 [2]

a) Evaluate: $\int \frac{dx}{5+4\cos x}$

Step 1: Use the substitution $t=\tan \frac{x}{2}$ We know that: $$\cos x=\frac{1-t^2}{1+t^2},dx=\frac{2}{1+t^2}dt$$ Substituting these into the integral: $$\int \frac{dx}{5+4\cos x}=\int \frac{\frac{2}{1+t^2}dt}{5+4\left(\frac{1-t^2}{1+t^2}\right)}$$

Step 2: Simplify the integrand Multiply numerator and denominator by $1+t^2$: $$=\int \frac{2dt}{5(1+t^2)+4(1-t^2)}=\int \frac{2dt}{5+5t^2+4-4t^2}=\int \frac{2dt}{9+t^2}$$

Step 3: Factor and integrate Rewrite the denominator: $$9+t^2=(3{)}^2+t^2$$ The integral becomes: $$2\int \frac{dt}{t^2+3^2}=2\cdot \frac{1}{3}{\tan }^{-1}\left(\frac{t}{3}\right)+C=\frac{2}{3}{\tan }^{-1}\left(\frac{t}{3}\right)+C$$

Step 4: Back-substitute $t=\tan \frac{x}{2}$ $$\boxed{\frac{2}{3}{\tan }^{-1}\left(\frac{\tan \frac{x}{2}}{3}\right)+C}$$

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b) Solve the differential equation $\sqrt{1-x^2}dy+\sqrt{1-y^2}dx=0$

Step 1: Rewrite the equation $$\sqrt{1-x^2}dy=-\sqrt{1-y^2}dx$$ Separate variables: $$\frac{dy}{\sqrt{1-y^2}}=-\frac{dx}{\sqrt{1-x^2}}$$

Step 2: Integrate both sides $$\int \frac{dy}{\sqrt{1-y^2}}=-\int \frac{dx}{\sqrt{1-x^2}}$$ The integrals are standard forms: $${\sin }^{-1}y=-{\sin }^{-1}x+C$$

Step 3: Solve for the general solution $${\sin }^{-1}y+{\sin }^{-1}x=C$$ Take the sine of both sides (using $\sin (A+B)=\sin A\cos B+\cos A\sin B$): $$y\sqrt{1-x^2}+x\sqrt{1-y^2}=\sin C$$ Let $\sin C=k$ (constant), so the implicit solution is: $$\boxed{y\sqrt{1-x^2}+x\sqrt{1-y^2}=k}$$