Question

Use substitution rule

1. Integral (cos x) e^sin(x) dx

Re write using the double angle formula
2. Integral cos^2 (x) dx

3. Integral cos square root x / square root x dx

4. Integral 1/ cube root (1-7x) dx

5. Integral 1/ 2+5x dx

Answer 1

1. The integral of
`\(\cos(x) e^(\sin(x)) dx\)` is
`\(e^(\sin(x)) + C\)`, where
`\(C\)` is the constant of integration.

2. The integral of
`\(\cos^2(x) dx\)` can be rewritten using the double-angle formula as
`\((x)/(2) + (\sin(2x))/(4) + C\).`

3. The integral of
`\((\cos(√(x)))/(√(x)) dx\)` is
`\(2\sin(√(x)) + C\)`.

4. The integral of
`\(\frac{1}{\sqrt[3]{1-7x}} dx\)`is
`\(-(3(1 - 7x)^(2/3))/(14) + C\)`.

5. The integral of
`\((1)/(2+5x) dx\)` is
`\((\ln|2+5x|)/(5) + C\)`.

Explanation:

Let's break down each integral:

1. Integral of
`\(\cos(x) e^(\sin(x)) dx\)`

This is a case of a simple substitution. We can let
`\(u = \sin(x)\)`, then
`\(du = \cos(x) dx\)`. Substituting these into the integral, we get
`\(\int e^u du\)`, which is simply
`\(e^u + C\)`. Substituting back for
`\(u\)`, we get
`\(e^(\sin(x)) + C\)`.

2. Integral of
`\(\cos^2(x) dx\)`

The double-angle formula is used here. We know that
`\(\cos^2(x) = (1 + \cos(2x))/(2)\)`. Substituting this into the integral, we get
`\(\int (1 + \cos(2x))/(2) dx\)`, which can be separated into two simpler integrals:
`\((1)/(2) \int dx + (1)/(2) \int \cos(2x) dx\)`. The integral of
`\(dx\) is \(x\)`, and the integral of
`\(\cos(2x)\)` is
`\((1)/(2)\sin(2x)\)`. So, the result is
`\((x)/(2) + (\sin(2x))/(4) + C\)`.

3. Integral of \(\frac{\cos(\sqrt{x})}{\sqrt{x}} dx\)

This is another case of simple substitution. We can let
`\(u = √(x)\)`, then
`\(du = (1)/(2√(x)) dx\), or \(2 du = (dx)/(√(x))\)`. Substituting these into the integral, we get
`\(2 \int \cos(u) du\), which is \(2\sin(u) + C\)`. Substituting back for
`\(u\)`, we get
`\(2\sin(√(x)) + C\)`.

4. Integral of
`\(\frac{1}{\sqrt[3]{1-7x}} dx\)`

Here, we can let
`\(u = 1 - 7x\)`, then
`\(du = -7 dx\)`, or
`\(-(1)/(7) du = dx\)`. Substituting these into the integral, we get
`\(-(1)/(7) \int u^(-1/3) du\)`, which is
`\(-(3)/(7)u^(2/3) + C\)`. Substituting back for
`\(u\)`, we get
`\(-(3(1 - 7x)^(2/3))/(14) + C\)`.

5. Integral of
`\((1)/(2+5x) dx\)`

This is a standard form of integral that results in a natural logarithm. The integral of
`\((1)/(a+bx) dx\)` is
`\((1)/(b) \ln |a+bx| + C\)`. So, the result is
`\((\ln|2+5x|)/(5) + C\)`.


Hope This Helps!