Calculus Help. Is wolfram alpha wrong?

Question

I have a question

I'm trying to evaluate the integral in wolfram alpha and its giving me a total different answer in the show step solution than the answer from my textbook
sin^4(3x)(dx)

In textbook:
Answer= -1/12sin^3(3x)cos(3x)-1/8sin(3x)cos(3x) +9/8(x) + C

Wolfram
Answer 1/96 (36 x-8 sin(6 x)+sin(12 x))+C


What gives? Is the Wolfram answer correct ?

Answer 1

Integral Calculator Wolfram

Answer 2

The two answers do not have to be equal. They only have to differ by a constant. Are you sure what you have as 9/8(x) should not be (3/8) x ?



In any case sin(6x) and sin(12x) can be expressed in terms of sin and cos of 3x via the double angle formulas, so check that also.

Answer 3

sin^4(3x) = ( sin^2(3x) ) ^2 = (1/2 ( 1 - cos(6x) )^2 = 1/4 ( 1 - 2 cos(6x) + cos^2(6x) )



= 1/4 ( 1 - 2 cos(6x) + 1/2( 1 + cos(12 x) )



Integrating, gives,



1/4 x - 1/12 sin(6x) + 1/8 x + (1/8)(1/12) sin(12x) + C



= 3/8 x - 1/12 sin(6x) + 1/96 sin(12 x) + c



= 1/96 ( 36 x - 8 sin(6x) + sin(12 x ) ) + C



So Wolfram Alpha solution is correct.

Answer 4

The answer could be written in several ways.

This is what you do. Assign a value for x. Calculate the expression you have using a calculator. Then, calculate the expression given by Wolframalpha. They should both be the same if your answer were correct.

For example, you may test using x=pi/6 or any other number.

Answer 5

nope, the two are equivalent