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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

Use substitution rule 1. Integral (cos x) e^sin(x) dx Re write using the double angle-example-1
User Ronash
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1 Answer

5 votes

Answer:

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!

User Michael Washington
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