what is the indefinite integral of (sin(4x))^4?

Question

anonymous · Answer

the answer is 3pi/32

anonymous · Answer

i dont know what to do with the 4x on th inside :S

anonymous · Answer

I think you're looking at the wrong answer (3pi/32). An indefinite integral is not a number; it's a function. A *definite* integral is a number.

anonymous · Answer

$$\int\limits \sin^{4}(4x)dx = \int\limits \sin^{2}(4x) \sin^{2}(4x) dx = \int\limits \sin^{2}(4x)(1-\cos^{2}(4x))dx$$

$$= \int\limits \sin^{2}(4x) dx - \int\limits \sin^{2}(4x)cos^{2}(4x) dx$$

$$= \frac{1}{2}\int\limits (1-\cos(8x))dx - \frac{1}{2}\int\limits \sin^{2}(4x)(1+\cos(8x))dx$$ 

$$= \frac{1}{2}\int\limits dx - \frac{1}{2} \int\limits \cos(8x)dx - \frac{1}{2}\int\limits \sin^{2}(4x)dx - \frac{1}{2}\int\limits \sin^{2}(4x)\cos(8x))dx  $$ 

$$= \frac{1}{2}x - \frac{1}{16}\sin(8x) - \frac{1}{4}\int\limits (1-\cos(8x))dx - \frac{1}{4}\int\limits (1 - cos(8x))\cos(8x))dx   $$  

$$= \frac{1}{2}x - \frac{1}{16}\sin(8x) - \frac{1}{4}x + \frac{1}{16}\sin(8x) - \frac{1}{4} \int\limits \cos(8x)dx + \frac{1}{4} \int\limits \cos^{2}(8x)dx$$

$$= \frac{1}{2}x - \frac{1}{16}\sin(8x) - \frac{1}{4}x + \frac{1}{16}\sin(8x) - \frac{1}{32} \sin(8x) + \frac{1}{8} \int\limits\limits (1-\sin(16x)dx $$

$$= \frac{1}{2}x - \frac{1}{16}\sin(8x) - \frac{1}{4}x + \frac{1}{16}\sin(8x) - \frac{1}{32} \sin(8x) + \frac{1}{8}x + \frac{1}{128}\cos(16x)  + C$$ 
Then simplify

anonymous · Answer

$$ = \frac{3}{8}x - \frac{1}{32}\sin(8x) + \frac{1}{128}\cos(16x) + C$$
Is what that simplifies to, though wolfram alpha got $$ = \frac{3}{8}x - \frac{1}{16}\sin(8x) + \frac{1}{128}\cos(16x) + C$$

So I may have messed up a fraction somewhere...

anonymous · Answer

ok, ssry i plugged in my numbers to solve after thats why lol

anonymous · Answer

how do you deal with the 4x? i dont get that part

turingtest · Answer

he used the identity$$\sin^2x=\frac12(1-\cos(2x))$$so$$\sin^2(4x)=\frac12(1-\cos(8x))$$