a) Evaluate: dx1 + x + x . [4] b) Solve: ^2 x dydx + y = 1 . [4] OR Solve: (1+x) dydx xy = 1 x .

a) Evaluate: $\int \frac{dx}{1+\sin x+\cos x}$

Step 1: Simplify the denominator

The integral is: $$I=\int \frac{dx}{1+\sin x+\cos x}$$ To simplify the denominator $1+\sin x+\cos x$, we use the following trigonometric identity: $$1+\sin x+\cos x=2\sin \left(\frac{x}{2}\right)\cos \left(\frac{x}{2}\right)+2{\cos }^2\left(\frac{x}{2}\right)$$ This comes from the half-angle identities: $$\sin x=2\sin \left(\frac{x}{2}\right)\cos \left(\frac{x}{2}\right),\cos x=2{\cos }^2\left(\frac{x}{2}\right)-1$$ Substituting these into the denominator: $$1+\sin x+\cos x=1+2\sin \left(\frac{x}{2}\right)\cos \left(\frac{x}{2}\right)+2{\cos }^2\left(\frac{x}{2}\right)-1$$ $$=2\sin \left(\frac{x}{2}\right)\cos \left(\frac{x}{2}\right)+2{\cos }^2\left(\frac{x}{2}\right)$$ $$=2\cos \left(\frac{x}{2}\right)\left[\sin \left(\frac{x}{2}\right)+\cos \left(\frac{x}{2}\right)\right]$$

Step 2: Rewrite the integral

Now, the integral becomes: $$I=\int \frac{dx}{2\cos \left(\frac{x}{2}\right)\left[\sin \left(\frac{x}{2}\right)+\cos \left(\frac{x}{2}\right)\right]}$$

Step 3: Use substitution

Let $t=\tan \left(\frac{x}{2}\right)$. Then: $$\sin \left(\frac{x}{2}\right)=\frac{t}{\sqrt{1+t^2}},\cos \left(\frac{x}{2}\right)=\frac{1}{\sqrt{1+t^2}},dx=\frac{2dt}{1+t^2}$$ Substituting these into the integral: $$I=\int \frac{\frac{2dt}{1+t^2}}{2\cdot \frac{1}{\sqrt{1+t^2}}\left[\frac{t}{\sqrt{1+t^2}}+\frac{1}{\sqrt{1+t^2}}\right]}$$ Simplify the denominator: $$2\cdot \frac{1}{\sqrt{1+t^2}}\left[\frac{t+1}{\sqrt{1+t^2}}\right]=\frac{2(t+1)}{1+t^2}$$ Thus, the integral becomes: $$I=\int \frac{2dt}{1+t^2}\cdot \frac{1+t^2}{2(t+1)}=\int \frac{dt}{t+1}$$

Step 4: Integrate

The integral simplifies to: $$I=\int \frac{dt}{t+1}=\ln |t+1|+C$$ Substituting back $t=\tan \left(\frac{x}{2}\right)$: $$I=\ln \left|\tan \left(\frac{x}{2}\right)+1\right|+C$$

Step 5: Further simplification (optional)

Using the identity $\tan \left(\frac{x}{2}\right)=\frac{1-\cos x}{\sin x}$, we can rewrite the answer as: $$I=\ln \left|\frac{1-\cos x+\sin x}{\sin x}\right|+C=\ln |1-\cos x+\sin x|-\ln |\sin x|+C$$ However, the simplest form is: $$\boxed{I=\ln \left|\tan \left(\frac{x}{2}\right)+1\right|+C}$$

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b) Solve: ${\cos }^{2}x\frac{dy}{dx}+y=1$

Step 1: Rewrite the differential equation

The given differential equation is: $${\cos }^{2}x\frac{dy}{dx}+y=1$$ This can be rewritten as: $$\frac{dy}{dx}+\frac{y}{{\cos }^{2}x}=\frac{1}{{\cos }^{2}x}$$

Step 2: Identify the type of differential equation

This is a linear first-order differential equation of the form: $$\frac{dy}{dx}+P(x)y=Q(x)$$ where $P(x)=\frac{1}{{\cos }^{2}x}$ and $Q(x)=\frac{1}{{\cos }^{2}x}$.

Step 3: Find the integrating factor (IF)

The integrating factor $\mu (x)$ is given by: $$\mu (x)=e^{\int P(x)dx}=e^{\int \frac{1}{{\cos }^{2}x}dx}=e^{\tan x}$$

Step 4: Multiply through by the integrating factor

Multiply both sides of the differential equation by $\mu (x)=e^{\tan x}$: $$e^{\tan x}\frac{dy}{dx}+e^{\tan x}\frac{y}{{\cos }^{2}x}=\frac{e^{\tan x}}{{\cos }^{2}x}$$ The left side is the derivative of $y{e}^{\tan x}$: $$\frac{d}{dx}\left(y{e}^{\tan x}\right)=\frac{e^{\tan x}}{{\cos }^{2}x}$$

Step 5: Integrate both sides

Integrate both sides with respect to $x$: $$y{e}^{\tan x}=\int \frac{e^{\tan x}}{{\cos }^{2}x}dx$$ Let $u=\tan x$, then $du=\frac{1}{{\cos }^{2}x}dx$. Thus: $$y{e}^{\tan x}=\int e^{u}du=e^u+C=e^{\tan x}+C$$

Step 6: Solve for $y$

Divide both sides by $e^{\tan x}$: $$y=1+C{e}^{-\tan x}$$

Final Answer:

$$\boxed{y=1+C{e}^{-\tan x}}$$

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OR

Solve: $(1+x)\frac{dy}{dx}-xy=1-x$

Step 1: Rewrite the differential equation

The given differential equation is: $$(1+x)\frac{dy}{dx}-xy=1-x$$ Divide both sides by $(1+x)$ to isolate $\frac{dy}{dx}$: $$\frac{dy}{dx}-\frac{x}{1+x}y=\frac{1-x}{1+x}$$

Step 2: Identify the type of differential equation

This is a linear first-order differential equation of the form: $$\frac{dy}{dx}+P(x)y=Q(x)$$ where $P(x)=-\frac{x}{1+x}$ and $Q(x)=\frac{1-x}{1+x}$.

Step 3: Find the integrating factor (IF)

The integrating factor $\mu (x)$ is given by: $$\mu (x)=e^{\int P(x)dx}=e^{\int -\frac{x}{1+x}dx}$$ Simplify the integrand: $$-\frac{x}{1+x}=-1+\frac{1}{1+x}$$ Thus: $$\int -\frac{x}{1+x}dx=\int \left(-1+\frac{1}{1+x}\right)dx=-x+\ln |1+x|$$ So the integrating factor is: $$\mu (x)=e^{-x+\ln |1+x|}=e^{-x}(1+x)$$

Step 4: Multiply through by the integrating factor

Multiply both sides of the differential equation by $\mu (x)=e^{-x}(1+x)$: $$e^{-x}(1+x)\frac{dy}{dx}-x{e}^{-x}y=(1-x)e^{-x}$$ The left side is the derivative of $y{e}^{-x}(1+x)$: $$\frac{d}{dx}\left(y{e}^{-x}(1+x)\right)=(1-x)e^{-x}$$

Step 5: Integrate both sides

Integrate both sides with respect to $x$: $$y{e}^{-x}(1+x)=\int (1-x)e^{-x}dx$$ To solve the integral on the right, use integration by parts. Let: $$u=1-x,dv=e^{-x}dx$$ Then: $$du=-dx,v=-e^{-x}$$ Thus: $$\int (1-x)e^{-x}dx=-(1-x)e^{-x}-\int e^{-x}dx=-(1-x)e^{-x}-(-e^{-x})+C=-(1-x)e^{-x}+e^{-x}+C$$ $$=e^{-x}(x)+C$$

Step 6: Solve for $y$

Substitute back: $$y{e}^{-x}(1+x)=e^{-x}(x)+C$$ Divide both sides by $e^{-x}(1+x)$: $$y=\frac{x}{1+x}+\frac{C}{e^{-x}(1+x)}=\frac{x}{1+x}+C{e}^x(1+x{)}^{-1}$$

Final Answer:

$$\boxed{y=\frac{x}{1+x}+C{e}^x(1+x{)}^{-1}}$$