What is fundamentally wrong with my approach?

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I've got a couple of these sort of questions wrong, and its pretty demoralising. Is it the method I'm using, or is it just arithmetic/silly errors? Here's one:

Use trigonometric identities before using a substitution (or reversing the chain rule) to integrate the following.

part c) $int sinleft(xright)cosleft(xright)e^cos2x$

Sub $u=cos(x)$

$int sinleft(xright)ue^cosleft(2xright)dx=int sinleft(xright)ue^2u^2-1dx:$

$=-1int ue^2u^2-1du:$

Then integration by parts:

$fracu2e^2u-1-int :frac12e^2u^2-1=fracu2e^2u-1-frac14e^2u^2-1$

$=fraccosleft(xright)2e^2cos^2x-1-frac14e^2cos^2x-1$

$-1cdot fraccosleft(xright)2e^2cos^2x-1-frac14e^2cos^2x-1=frac-cosleft(xright)2e^2cos^2x-1+frac14e^2cos^2x-1$

integration trigonometry proof-verification

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edited Jul 28 at 23:29

asked Jul 28 at 23:08

Cheks Nweze

627

• 
  
  
  

  
  4
  

  
  
  
  
  
  

  
  The simpler way is $sin x cos x = sin 2x / 2$ (literally before using a substitution...).
  – metamorphy
  Jul 28 at 23:11
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  I can't believe I didn't think of that. I just tried it and got the right answer that way (thanks!). I wonder what was wrong with my more long winded approach, though.
  – Cheks Nweze
  Jul 28 at 23:20
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Where you write $int sinleft(xright)ue^2u^2-1dx = -1int ue^2u-1du$ Have you accidentally dropped $u^2$ in the exponent to $u$?
  – Matt
  Jul 28 at 23:26
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  The question states "use trigonometric identities before using a substitution" and the first thing you did was use a substitution!
  – steven gregory
  Jul 28 at 23:35
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Your integration by parts is indeed wrong, as promised :). Write out $f$ and $g$ in $int f'(u) g(u) du$, you'll see the problem.
  – Maxim
  Jul 29 at 9:59
  

  
  
  
  

 | 
show 1 more comment

up vote
1
down vote

favorite

I've got a couple of these sort of questions wrong, and its pretty demoralising. Is it the method I'm using, or is it just arithmetic/silly errors? Here's one:

Use trigonometric identities before using a substitution (or reversing the chain rule) to integrate the following.

part c) $int sinleft(xright)cosleft(xright)e^cos2x$

Sub $u=cos(x)$

$int sinleft(xright)ue^cosleft(2xright)dx=int sinleft(xright)ue^2u^2-1dx:$

$=-1int ue^2u^2-1du:$

Then integration by parts:

$fracu2e^2u-1-int :frac12e^2u^2-1=fracu2e^2u-1-frac14e^2u^2-1$

$=fraccosleft(xright)2e^2cos^2x-1-frac14e^2cos^2x-1$

$-1cdot fraccosleft(xright)2e^2cos^2x-1-frac14e^2cos^2x-1=frac-cosleft(xright)2e^2cos^2x-1+frac14e^2cos^2x-1$

integration trigonometry proof-verification

share|cite|improve this question

edited Jul 28 at 23:29

asked Jul 28 at 23:08

Cheks Nweze

627

• 
  
  
  

  
  4
  

  
  
  
  
  
  

  
  The simpler way is $sin x cos x = sin 2x / 2$ (literally before using a substitution...).
  – metamorphy
  Jul 28 at 23:11
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  I can't believe I didn't think of that. I just tried it and got the right answer that way (thanks!). I wonder what was wrong with my more long winded approach, though.
  – Cheks Nweze
  Jul 28 at 23:20
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Where you write $int sinleft(xright)ue^2u^2-1dx = -1int ue^2u-1du$ Have you accidentally dropped $u^2$ in the exponent to $u$?
  – Matt
  Jul 28 at 23:26
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  The question states "use trigonometric identities before using a substitution" and the first thing you did was use a substitution!
  – steven gregory
  Jul 28 at 23:35
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Your integration by parts is indeed wrong, as promised :). Write out $f$ and $g$ in $int f'(u) g(u) du$, you'll see the problem.
  – Maxim
  Jul 29 at 9:59
  

  
  
  
  

 | 
show 1 more comment

up vote
1
down vote

favorite

up vote
1
down vote

favorite

I've got a couple of these sort of questions wrong, and its pretty demoralising. Is it the method I'm using, or is it just arithmetic/silly errors? Here's one:

Use trigonometric identities before using a substitution (or reversing the chain rule) to integrate the following.

part c) $int sinleft(xright)cosleft(xright)e^cos2x$

Sub $u=cos(x)$

$int sinleft(xright)ue^cosleft(2xright)dx=int sinleft(xright)ue^2u^2-1dx:$

$=-1int ue^2u^2-1du:$

Then integration by parts:

$fracu2e^2u-1-int :frac12e^2u^2-1=fracu2e^2u-1-frac14e^2u^2-1$

$=fraccosleft(xright)2e^2cos^2x-1-frac14e^2cos^2x-1$

$-1cdot fraccosleft(xright)2e^2cos^2x-1-frac14e^2cos^2x-1=frac-cosleft(xright)2e^2cos^2x-1+frac14e^2cos^2x-1$

integration trigonometry proof-verification

share|cite|improve this question

edited Jul 28 at 23:29

asked Jul 28 at 23:08

Cheks Nweze

627

I've got a couple of these sort of questions wrong, and its pretty demoralising. Is it the method I'm using, or is it just arithmetic/silly errors? Here's one:

Use trigonometric identities before using a substitution (or reversing the chain rule) to integrate the following.

part c) $int sinleft(xright)cosleft(xright)e^cos2x$

Sub $u=cos(x)$

$int sinleft(xright)ue^cosleft(2xright)dx=int sinleft(xright)ue^2u^2-1dx:$

$=-1int ue^2u^2-1du:$

Then integration by parts:

$fracu2e^2u-1-int :frac12e^2u^2-1=fracu2e^2u-1-frac14e^2u^2-1$

$=fraccosleft(xright)2e^2cos^2x-1-frac14e^2cos^2x-1$

$-1cdot fraccosleft(xright)2e^2cos^2x-1-frac14e^2cos^2x-1=frac-cosleft(xright)2e^2cos^2x-1+frac14e^2cos^2x-1$

integration trigonometry proof-verification

share|cite|improve this question

edited Jul 28 at 23:29

asked Jul 28 at 23:08

Cheks Nweze

627

share|cite|improve this question

share|cite|improve this question

edited Jul 28 at 23:29

edited Jul 28 at 23:29

edited Jul 28 at 23:29

asked Jul 28 at 23:08

Cheks Nweze

627

asked Jul 28 at 23:08

Cheks Nweze

627

asked Jul 28 at 23:08

Cheks Nweze

627

627

• 
  
  
  

  
  4
  

  
  
  
  
  
  

  
  The simpler way is $sin x cos x = sin 2x / 2$ (literally before using a substitution...).
  – metamorphy
  Jul 28 at 23:11
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  I can't believe I didn't think of that. I just tried it and got the right answer that way (thanks!). I wonder what was wrong with my more long winded approach, though.
  – Cheks Nweze
  Jul 28 at 23:20
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Where you write $int sinleft(xright)ue^2u^2-1dx = -1int ue^2u-1du$ Have you accidentally dropped $u^2$ in the exponent to $u$?
  – Matt
  Jul 28 at 23:26
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  The question states "use trigonometric identities before using a substitution" and the first thing you did was use a substitution!
  – steven gregory
  Jul 28 at 23:35
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Your integration by parts is indeed wrong, as promised :). Write out $f$ and $g$ in $int f'(u) g(u) du$, you'll see the problem.
  – Maxim
  Jul 29 at 9:59
  

  
  
  
  

 | 
show 1 more comment

• 
  
  
  

  
  4
  

  
  
  
  
  
  

  
  The simpler way is $sin x cos x = sin 2x / 2$ (literally before using a substitution...).
  – metamorphy
  Jul 28 at 23:11
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  I can't believe I didn't think of that. I just tried it and got the right answer that way (thanks!). I wonder what was wrong with my more long winded approach, though.
  – Cheks Nweze
  Jul 28 at 23:20
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Where you write $int sinleft(xright)ue^2u^2-1dx = -1int ue^2u-1du$ Have you accidentally dropped $u^2$ in the exponent to $u$?
  – Matt
  Jul 28 at 23:26
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  The question states "use trigonometric identities before using a substitution" and the first thing you did was use a substitution!
  – steven gregory
  Jul 28 at 23:35
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Your integration by parts is indeed wrong, as promised :). Write out $f$ and $g$ in $int f'(u) g(u) du$, you'll see the problem.
  – Maxim
  Jul 29 at 9:59
  

  
  
  
  

4

4

The simpler way is $sin x cos x = sin 2x / 2$ (literally before using a substitution...).
– metamorphy
Jul 28 at 23:11

The simpler way is $sin x cos x = sin 2x / 2$ (literally before using a substitution...).
– metamorphy
Jul 28 at 23:11

I can't believe I didn't think of that. I just tried it and got the right answer that way (thanks!). I wonder what was wrong with my more long winded approach, though.
– Cheks Nweze
Jul 28 at 23:20

I can't believe I didn't think of that. I just tried it and got the right answer that way (thanks!). I wonder what was wrong with my more long winded approach, though.
– Cheks Nweze
Jul 28 at 23:20

1

1

Where you write $int sinleft(xright)ue^2u^2-1dx = -1int ue^2u-1du$ Have you accidentally dropped $u^2$ in the exponent to $u$?
– Matt
Jul 28 at 23:26

Where you write $int sinleft(xright)ue^2u^2-1dx = -1int ue^2u-1du$ Have you accidentally dropped $u^2$ in the exponent to $u$?
– Matt
Jul 28 at 23:26

1

1

The question states "use trigonometric identities before using a substitution" and the first thing you did was use a substitution!
– steven gregory
Jul 28 at 23:35

The question states "use trigonometric identities before using a substitution" and the first thing you did was use a substitution!
– steven gregory
Jul 28 at 23:35

1

1

Your integration by parts is indeed wrong, as promised :). Write out $f$ and $g$ in $int f'(u) g(u) du$, you'll see the problem.
– Maxim
Jul 29 at 9:59

Your integration by parts is indeed wrong, as promised :). Write out $f$ and $g$ in $int f'(u) g(u) du$, you'll see the problem.
– Maxim
Jul 29 at 9:59

 | 
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