W02 Regular

Due date: Sunday 1/25, 11:59pm

Trig power products

01

04

All odd power product

Compute the integral:

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Solution

Solutions - 0040-04

(1) Notice odd power on . Swap the even bunch:

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(2) Perform -sub setting and thus :

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(3) Convert back to :

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02

05

Tangent and secant mixed parity

Compute the integral:

(a) Using .

(b) Using .

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Solution

Solutions - 0040-05

(a) Select and thus :

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(b)

(1) Select and thus :

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(3) Swap even bunch using :

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(4) Perform -sub with and integrate:

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03

06

Power product with negative power

Compute the integral:

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Solution

Solutions - 0040-06

(1) Change variable by substituting and :

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(2) Identify :

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(3) Perform -sub with and thus :

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Trig substitution

04

03

Trig sub

Compute the integral:

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Solution

Solutions - 0050-03

(1) Notice pattern, so we should make use of the identity .

Select and thus . Then:

Plug in and simplify:

(We must assume that for the relevant values of here.)

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(2) Use power-to-frequency conversion:

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(3) Convert back to terms of :

First draw a triangle expressing :

Therefore:

For , use the double-angle identity:

Therefore:

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05

04

Trig sub

Compute the integral:

Hint: complete the square and then substitute.

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Solution

Solutions - 0050-04

(1) Complete the square:

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(2) Substitute and thus :

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(3) Convert back to terms of :

First draw a triangle expressing :

It follows that . Then:

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Note A: Using log rules, the denominator can be brought out as which can be “absorbed” into the constant .

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06

05

Trig sub

Compute the integral:

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Solution

Solutions - 0050-05

(1) Notice pattern, so we should make use of the identity .

Select and thus . Then:

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(2) Convert to and integrate:

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(3) Convert back to terms of :

Draw a triangle expressing :

Therefore and . Then:

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07

06

Double sub: -sub then trig sub

Compute the definite integral:

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Solution

Solutions - 0050-06

(1) Perform -sub setting and thus . Adjust the bounds as follows:

Therefore:

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(2) Notice pattern, so we should make use of the identity .

Select and thus . Adjust bounds:

Therefore:

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(3) Integrate from memory or multiplying above and below by :

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08

07

Trig sub for electric charge

A charged wire lies on the -axis running from to . The electric field at the point is given by:

Find the numerical value of assuming and and and .

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Solution

Solutions - 0050-07

(1) Take out constants and insert given values:

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(2) Notice pattern, so we should make use of the identity .

Select and thus . Then:

Adjust bounds:

Then:

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(3) Integrate:

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(4) Compute :

Draw a triangle expressing :

Therefore . Then:

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