# Major Integration Help

• Jan 13th 2009, 06:24 PM
dillonmhudson
Major Integration Help
A lot of question on some really hard homework I have. Any help would be greatly appreciated. Thanks.

1. int [ (x^2 - 3x + 2) / (x+1) * dx ]

2. int [ x^(1/2) / (x^(1/2) - 3) * dx ]

3. int [ tan 5x * dx ]
___An even problem, is the answer:
___-ln | csc(2x) + cot(2x) | + c

4. int [ sec (x/2) * dx ]
___An even problem, is the answer:
___2*ln | sec(x/2) + tan(x/2) | + c

5. int 2/0 [ (x^2 - 2) / (x + 1) * dx ]

Thanks Again!
• Jan 13th 2009, 06:46 PM
Prove It
Quote:

Originally Posted by dillonmhudson
A lot of question on some really hard homework I have. Any help would be greatly appreciated. Thanks.

1. int [ (x^2 - 3x + 2) / (x+1) * dx ]

2. int [ x^(1/2) / (x^(1/2) - 3) * dx ]

3. int [ tan 5x * dx ]
___An even problem, is the answer:
___-ln | csc(2x) + cot(2x) | + c

4. int [ sec (x/2) * dx ]
___An even problem, is the answer:
___2*ln | sec(x/2) + tan(x/2) | + c

5. int 2/0 [ (x^2 - 2) / (x + 1) * dx ]

Thanks Again!

1. Are you sure the numerator isn't $x^2 + 3x + 2$???
• Jan 13th 2009, 06:50 PM
Prove It
Quote:

Originally Posted by dillonmhudson
A lot of question on some really hard homework I have. Any help would be greatly appreciated. Thanks.

1. int [ (x^2 - 3x + 2) / (x+1) * dx ]

2. int [ x^(1/2) / (x^(1/2) - 3) * dx ]

3. int [ tan 5x * dx ]
___An even problem, is the answer:
___-ln | csc(2x) + cot(2x) | + c

4. int [ sec (x/2) * dx ]
___An even problem, is the answer:
___2*ln | sec(x/2) + tan(x/2) | + c

5. int 2/0 [ (x^2 - 2) / (x + 1) * dx ]

Thanks Again!

1. Long divide. Then the integrand becomes $x - 4 - \frac{2}{x + 1}$. Much easier.
• Jan 13th 2009, 06:51 PM
dillonmhudson
You thinking about factoring? Yeah I'm positive, I double checked that.
• Jan 13th 2009, 06:52 PM
Prove It
Quote:

Originally Posted by dillonmhudson
You thinking about factoring? Yeah I'm positive, I double checked that.

Worth a shot.

You'll have to long divide in that case.
• Jan 13th 2009, 06:54 PM
Prove It
Quote:

Originally Posted by dillonmhudson
A lot of question on some really hard homework I have. Any help would be greatly appreciated. Thanks.

1. int [ (x^2 - 3x + 2) / (x+1) * dx ]

2. int [ x^(1/2) / (x^(1/2) - 3) * dx ]

3. int [ tan 5x * dx ]
___An even problem, is the answer:
___-ln | csc(2x) + cot(2x) | + c

4. int [ sec (x/2) * dx ]
___An even problem, is the answer:
___2*ln | sec(x/2) + tan(x/2) | + c

5. int 2/0 [ (x^2 - 2) / (x + 1) * dx ]

Thanks Again!

3. Remember that $\tan{5x} = \frac{\sin{5x}}{\cos{5x}}$.

Make the substitution $u = \cos{5x}$.
• Jan 13th 2009, 06:56 PM
Prove It
Quote:

Originally Posted by dillonmhudson
A lot of question on some really hard homework I have. Any help would be greatly appreciated. Thanks.

1. int [ (x^2 - 3x + 2) / (x+1) * dx ]

2. int [ x^(1/2) / (x^(1/2) - 3) * dx ]

3. int [ tan 5x * dx ]
___An even problem, is the answer:
___-ln | csc(2x) + cot(2x) | + c

4. int [ sec (x/2) * dx ]
___An even problem, is the answer:
___2*ln | sec(x/2) + tan(x/2) | + c

5. int 2/0 [ (x^2 - 2) / (x + 1) * dx ]

Thanks Again!

For 5. You'll have to long divide as well.

$\frac{x^2 - 1}{x + 1} = x - 1 - \frac{1}{x + 1}$.
• Jan 13th 2009, 07:00 PM
DeMath
Quote:

Originally Posted by dillonmhudson
A lot of question on some really hard homework I have. Any help would be greatly appreciated. Thanks.

1. int [ (x^2 - 3x + 2) / (x+1) * dx ]

2. int [ x^(1/2) / (x^(1/2) - 3) * dx ]

3. int [ tan 5x * dx ]
___An even problem, is the answer:
___-ln | csc(2x) + cot(2x) | + c

4. int [ sec (x/2) * dx ]
___An even problem, is the answer:
___2*ln | sec(x/2) + tan(x/2) | + c

5. int 2/0 [ (x^2 - 2) / (x + 1) * dx ]

Thanks Again!

$1.{\text{ }}\int {\frac{{{x^2} - 3x + 2}}{{x + 1}}} dx = \int {\frac{{\left( {x + 1} \right)x - \left( {4x - 2} \right)}}{{x + 1}}} dx = \int {\left[ {x - \frac{{4x - 2}}{{x + 1}}} \right]} dx =$

$= \frac{{{x^2}}}{2} - 2\int {\frac{{2x - 1}}{{x + 1}}} dx = \frac{{{x^2}}}{2} - 2\int {\frac{{2x + 2 - 3}}{{x + 1}}} dx = \frac{{{x^2}}}
{2} - 2\int {\left[ {2 - \frac{3}{{x + 1}}} \right]} dx =$

$= \frac{{{x^2}}}{2} - 4\int {dx} + 6\int {\frac{{dx}}{{x + 1}}} = \frac{{{x^2}}}{2} - 4x + 6\ln \left( {x + 1} \right) + C.$
• Jan 14th 2009, 11:21 AM
Krizalid
Quote:

Originally Posted by DeMath

$= \frac{{{x^2}}}{2} - 4\int {dx} + 6\int {\frac{{dx}}{{x + 1}}} = \frac{{{x^2}}}{2} - 4x + 6\ln \left( {x + 1} \right) + C.$

Not quite on the $\ln(x+1),$ something is missing... :eek:
• Jan 14th 2009, 11:39 AM
DeMath
This $\ln \left| {x + 1} \right|$?
(Clapping)
Thanks! (Bow)
• Jan 14th 2009, 12:47 PM
stevedave
Quote:

Originally Posted by Prove It
1. Long divide. Then the integrand becomes $x - 4 - \frac{2}{x + 1}$. Much easier.

I think the integrand should be $x - 4 + \frac{6}{x + 1}$ after the long division.