Monday

September 1, 2014

September 1, 2014

Posted by **Rox** on Thursday, June 14, 2012 at 10:37pm.

find the integral

- calculus -
**Steve**, Friday, June 15, 2012 at 10:42amone way:

sin(3x) = sinx cos2x + cosx sin2x

= sinx(2cos^2(x)-1) + 2cosx sinx cosx

= cos^2(x) sinx - sinx + 2cos^2(x) sinx

= 3cos^2(x)sinx - sinx

∫ = -cos^3(x) + cos(x)

another way:

sinx sin3x = 1/2 cos(2x) - 1/2 cos(4x)

∫ = 1/4 sin2x - 1/8 sin4x

a little manipulation of that also yields

sin^3(x) - cos(x)

the various expressions are not identical, but differ only by a constant C

**Answer this Question**

**Related Questions**

Calculus - Hello, I just wanted to verify if my work was good. Calculate the ...

Definite integral by parts (correction) - Hello, I just wanted to verify if my ...

Calculus AP - Evaluate the integral interval from [0 to pi] t sin(3t)dt Use ...

Math integrals - What is the indefinite integral of ∫ [sin (π/x)]/ x^...

Integral - That's the same as the integral of sin^2 x dx. Use integration by ...

calc - find integral using table of integrals ) integral sin^4xdx this the ...

Calculus AP - hi again im really need help TextBook: James Stewart:Essential ...

Calculus - Evaluate the integral. S= integral sign I= absolute value S ((cos x...

math - I'm trying to find the convolution f*g where f(t)=g(t)=sin(t). I set up ...

Integral Help - 1.) ∫ (sin x) / (cos^2 x) dx 2.) ∫ (1) / (1+x^2) dx ...