Substitute \(\displaystyle u = 7x\). Then \(\displaystyle \frac{\mathrm{d} u}{\mathrm{d} x} = 7\) and \(\displaystyle \mathrm{d} u = 7\: \mathrm{d} x\), or \(\displaystyle \mathrm{d} x = \frac{1}{7} \mathrm{d} u\).

The integral becomes

\(\displaystyle \int \sin 7x \; \mathrm{d} x\)

\(\displaystyle = \int \sin u \cdot \frac{1}{7}\; \mathrm{d} u\)

\(\displaystyle = \frac{1}{7} \int \sin u \; \mathrm{d} u\)

\(\displaystyle = \frac{1}{7} (- \cos u + C)\)

\(\displaystyle = - \frac{1}{7} \cos 7x + \frac{1}{7}C = - \frac{1}{7} \cos 7x + C\)

Note that the \(\displaystyle \frac{1} {7 }\) gets absorbed into the constant term.

As a separable differential equation, this can be solved as follows:

\(\displaystyle \frac{\mathrm{d} y}{\mathrm{d} x} = 2xy\)

\(\displaystyle \frac{\frac{\mathrm{d} y}{\mathrm{d} x} \cdot \mathrm{d} x }{y}= \frac{2xy \cdot \mathrm{d} x}{y}\)

\(\displaystyle \frac{\mathrm{d} y }{y}= 2x \; \mathrm{d} x\)

Now, integrate both sides:

\(\displaystyle \int \frac{\mathrm{d} y }{y}=\int 2x \; \mathrm{d} x\)

\(\displaystyle \ln y = x^2 + C\)

\(\displaystyle e^ { \ln y } = e^ { x^2 + C}\)

\(\displaystyle y= e^ { x^2 + C}\)

\(\displaystyle y= e^ { C} \cdot e^ { x^2 }\)

\(\displaystyle y= C_{0} e^ { x^2 }\)

Since the initial condition is \(\displaystyle y(1)= 2\), we can find \(\displaystyle C_{0}\) by substituting:

\(\displaystyle 2= C_{0} e^ { 1^2 }\)

\(\displaystyle 2= C_{0} e\)

\(\displaystyle \frac{2}{e}= \frac{C_{0} e}{e} = C_{0}\)

The solution is:

\(\displaystyle y=\frac{2}{e}\cdot e^ { x^2 }\)

\(\displaystyle y=2 \cdot \frac{e^ { x^2 }}{e}\)

\(\displaystyle y=2e^ { x^2 -1 }\)

As a separable differential equation, 

\(\displaystyle \frac{\mathrm{d} y}{\mathrm{d} x} = \sec^2 x \csc y\)

can be rewritten and solved as follows:

\(\displaystyle \frac{\mathrm{d} y}{\mathrm{d} x} = \sec^2 x \cdot \frac{1}{\sin y }\)

\(\displaystyle \frac{\mathrm{d} y}{\mathrm{d} x} \cdot \sin y \cdot \mathrm{d} x= \sec^2 x \cdot \frac{1}{\sin y } \cdot \sin y \cdot \mathrm{d} x\)

\(\displaystyle \sin y \; \mathrm{d} y = \sec^2 x \; \mathrm{d} x\)

\(\displaystyle \int \sin y \; \mathrm{d} y = \int \sec^2 x \; \mathrm{d} x\)

\(\displaystyle -\cos y = \tan x + C\)

\(\displaystyle \cos y = - \tan x - C\)

To find \(\displaystyle C\), use the initial condition \(\displaystyle y (0)= \frac{\pi }{6}\):

\(\displaystyle \cos \frac{\pi }{6} = - \tan 0 - C\)

\(\displaystyle \frac{ \sqrt{3}}{2} = 0 - C\)

\(\displaystyle C = -\frac{ \sqrt{3}}{2}\), so

\(\displaystyle \cos y = - \tan x + \frac{\sqrt{3}}{2}\)

and 

\(\displaystyle y =\cos ^{-1} \left ( - \tan x + \frac{\sqrt{3}}{2} \right )\)

As a separable differential equation, this can be solved as follows:

\(\displaystyle \frac{\mathrm{d} y}{\mathrm{d} x} = ye^{x}\)

\(\displaystyle \frac{\mathrm{d} y}{\mathrm{d} x} \cdot \frac{\mathrm{d} x}{y} = ye^{x} \cdot \frac{\mathrm{d} x}{y}\)

\(\displaystyle \frac{\mathrm{d} y} {y} = e^{x} \mathrm{d} x\)

\(\displaystyle \int \frac{\mathrm{d} y} {y} = \int e^{x} \mathrm{d} x\)

\(\displaystyle \ln y = e^{x} + C\)

\(\displaystyle e^ {\ln y} = e^{ e^{x}+ C}\)

\(\displaystyle y =e^{ C} \cdot e^{ e^{x}}\)

\(\displaystyle y =C_{0}\cdot e^{ e^{x}}\)

To find \(\displaystyle C_{0}\), substitute the initial conditions:

\(\displaystyle 10=C_{0}\cdot e^{ e^{0}}\)

\(\displaystyle 10=C_{0}\cdot e^{ 1}\)

\(\displaystyle 10=C_{0}\cdot e\)

\(\displaystyle \frac{10}{e}=C_{0}\)

The solution is \(\displaystyle y =\frac{10}{e}\cdot e^{ e^{x}} = \frac{10e^{ e^{x}}} {e} = 10e^{ e^{x}- 1}\).

As a separable differential equation, this can be solved as follows:

\(\displaystyle \frac{\mathrm{d} y}{\mathrm{d} x} = \sec x \csc y\)

\(\displaystyle \frac{\mathrm{d} y}{\mathrm{d} x} \cdot\frac{ \mathrm{d} x }{ \csc y}= \sec x \csc y \cdot\frac{ \mathrm{d} x }{ \csc y}\)

\(\displaystyle \frac{ \mathrm{d} y }{ \csc y}= \sec x \; \mathrm{d} x\)

\(\displaystyle \sin y \; \mathrm{d} y = \sec x \; \mathrm{d} x\)

\(\displaystyle \int \sin y \; \mathrm{d} y = \int \sec x \; \mathrm{d} x\)

\(\displaystyle -\cos y = \ln | \sec x + \tan x | + C\)

\(\displaystyle \cos y = - \ln | \sec x + \tan x | + C\)

\(\displaystyle y = \cos^{-1} \left (- \ln | \sec x + \tan x | + C \right )\)

Apply the initial condition to find \(\displaystyle C\):

\(\displaystyle -\frac{\pi}{2} = \cos^{-1} \left (- \ln | \sec 0 + \tan 0| + C \right )\)

\(\displaystyle -\frac{\pi}{2} = \cos^{-1} \left (- \ln |1 +0 + C | \right )\)

\(\displaystyle \cos \left (-\frac{\pi}{2} \right )= - \ln |1 +0 + C |\)

\(\displaystyle 0= - \ln |C +1 |\)

\(\displaystyle 0= \ln |C +1 |\)

\(\displaystyle e^0=e ^ { \ln |C +1 |}\)

\(\displaystyle 1 = C + 1\)

\(\displaystyle C= 0\)

The solution is

\(\displaystyle y = \cos^{-1} \left (- \ln | \sec x + \tan x | \right )\).

As a separable differential equation, 

\(\displaystyle \frac{\mathrm{d} y}{\mathrm{d} x} = xy + y\)

can be rewritten as follows:

\(\displaystyle \frac{\mathrm{d} y}{\mathrm{d} x} = \left (x+ 1 \right ) y\)

\(\displaystyle \frac{\mathrm{d} y}{\mathrm{d} x} \cdot \frac{\mathrm{d} x}{y} = \left (x+ 1 \right ) y \cdot \frac{\mathrm{d} x}{y}\)

\(\displaystyle \frac{\mathrm{d} y}{y} = \left (x+ 1 \right ) \mathrm{d} x\)

\(\displaystyle \int \frac{\mathrm{d} y}{y} = \int \left (x+ 1 \right ) \mathrm{d} x\)

\(\displaystyle \ln y = \frac{x^2}{2} + x + C\)

\(\displaystyle e^ {\ln y }= e^ { \frac{x^2}{2} + x + C}\)

\(\displaystyle y=C_{0} e^ { \frac{x^2}{2} + x }\)

We can find \(\displaystyle C_{0}\) using the initial condition \(\displaystyle y (0) = 3\):

\(\displaystyle 3=C_{0} e^ { \frac{0^2}{2} + 0 } = C_{0} e^ { 0 } = C_{0} \cdot 1 = C_{0}\)

The solution is 

\(\displaystyle y=3 e^ { \frac{x^2}{2} + x }\).

Integration by parts is the best strategy here.

Let \(\displaystyle u = x\) and \(\displaystyle \mathrm {d}v = e^{3x} \; \mathrm {d}x\). 

Then

\(\displaystyle \frac{\mathrm{d} u}{\mathrm{d} x} = 1\) and \(\displaystyle \mathrm{d} u = \mathrm{d} x\).

Also,

\(\displaystyle \mathrm {d}v = e^{3x} \; \mathrm {d}x\).

\(\displaystyle \int \mathrm {d}v = \int e^{3x} \; \mathrm {d}x\)

\(\displaystyle v = \frac{e^{3x}}{3} = \frac{1}{3}e^{3x}\)

Therefore,

\(\displaystyle \int xe^{3x} \;\mathrm{d}x\)

\(\displaystyle = \int u \; \mathrm{d}v\)

\(\displaystyle = uv- \int v \; \mathrm{d}u\)

\(\displaystyle = x \cdot \frac{1}{3}e^{3x}- \int \frac{1}{3}e^{3x} \cdot \; \mathrm{d}x\)

\(\displaystyle = \frac{1}{3}xe^{3x}- \frac{1}{3} \int e^{3x} \mathrm{d}x\)

\(\displaystyle = \frac{1}{3}xe^{3x}- \frac{1}{3}\cdot \left ( \frac{e^{3x}}{3} + C \right )\)

\(\displaystyle = \frac{1}{3}xe^{3x}- \frac{1}{9} e^{3x} + C\)

\(\displaystyle = \frac{3xe^{3x}-e^{3x}}{9} + C\)

Note the absorption of the negative into the constant in the third to last step.

As a separable differential equation, 

\(\displaystyle \frac{\mathrm{d} y}{\mathrm{d} x} = 2xye^{x^2}\)

can be rewritten and solved as follows:

\(\displaystyle \frac{\mathrm{d} y}{\mathrm{d} x} \cdot \frac{\mathrm{d} x}{y}= 2xye^{x^2}\cdot \frac{\mathrm{d} x}{y}\)

\(\displaystyle \frac{\mathrm{d} y}{y} = 2xe^{x^2}\mathrm{d} x\)

\(\displaystyle \int \frac{\mathrm{d} y}{y} = \int 2xe^{x^2}\mathrm{d} x\)

The integral on the right can be solved by setting \(\displaystyle u = x^2\) and, subsequently, \(\displaystyle \frac{\mathrm{d} u}{\mathrm{d} x} = \frac{\mathrm{d} }{\mathrm{d} x} \; x^2 = 2x\) and \(\displaystyle \mathrm{d} u = 2x \; \mathrm{d} x\).

\(\displaystyle \int \frac{\mathrm{d} y}{y} = \int e^{u}\mathrm{d} u\)

\(\displaystyle \ln y = e^u + C\)

\(\displaystyle \ln y = e^{x^2} + C\)

\(\displaystyle y = e^ { \e^{x^2} + C }\)

\(\displaystyle y = C_{0} e^ { e^{x^2} }\)

To find \(\displaystyle C_{0}\), use the initial conditions:

\(\displaystyle e^2= C_{0} e^ { e^{0^2} }\)

\(\displaystyle e^2= C_{0} e^ { e^{0} }\)

\(\displaystyle e^2= C_{0} e^ { 1}\)

\(\displaystyle e^2= C_{0} e\)

\(\displaystyle C_{0} = \frac{ e^2 }{e} = e\)

The solution is 

\(\displaystyle y = e \cdot e^ { e^{x^2} } = e^ { e^{x^2}+1 }\).

To integrate this function, use u substitution. Make,

 \(\displaystyle \\u=x^2+1 \\du=2xdx\) 

then substitute them into the equation to get

 \(\displaystyle f(x)=\frac{du}{u}\).

The integral of

 \(\displaystyle \frac{du}{u}=ln|u|+C\)

then plug u back into the equation

\(\displaystyle ln|x^2+1|+C\).

The +C is essential because the integral is indefinite.

To integrate this function, use u substitution. Make

 \(\displaystyle \\u=sec(x) \\du=sec(x)tan(x)dx\) 

then substitute them into the equation to get

 \(\displaystyle \int udu\).

The integral of

 \(\displaystyle udu=\frac{u^2}{2}+C\)

then plug u back into the equation

\(\displaystyle \frac{sec^2(x)}{2}+C\).

The +C is essential because the integral is indefinite.