Using U – Substitution or Integration by Parts, evaluate the Indefinite Integral of the following. ∫ sin x sin 4x dx dx Hint. Use the integration by parts twice.

Question

Using U – Substitution or Integration by Parts, evaluate the Indefinite

Integral of the following.

∫ sin x sin 4x dx dx Hint. Use the integration by parts twice.

Answer 1

In a way I sort of concur with answer given by Raymond.

isn't sinxsin4x=-(1/2)(cos5x-cos3x),and the integral computes to

I(x)=-(1/2)(((sin5x)/5)-((sin3x)/3))+C

back to integrating by parts

u=sin4x,,,,, dv=sinxdx

du=4cos4x,,,,,v=-cosx

I(x)=-cosxsin4x+∫ 4cos4xcosxdx

u=4cos4x ,,,,dv=cosxdx

du=-16sin4x,,,,,v=sinx

I(x)=-cosxsin4x+4sinxcos4x+16∫sinxsin4xdx

I(x)=-cosxsin4x+4sinxcos4x+16*I(x)

I(x)=(1/15)(cosxsin4x-4sinxcos4x)+C

Answer 2

integral of sinxsin4xdx

= sin^3(x)/3 - sin^5(x)/8 + C

use u=sinx

du = cosxdx

cos^2(x) = 1-sin^2(x)

sin4x = 8sinxcos^3(x) - 4sinxcosx

sinx(sin4x)dx = sinx(8sinxcos^3(x) -4sinxcosx)du/cosx

= u(8ucos^2(x) - 4u)du

=4u^2(2cos^2(x) -1)du

= 4u^2(2(1-sin^2(x)) -1)du

= 4u^2(2- 2u^2-1)du

= 4u^2(1-2u^2)du

= 4u^2 -8u^4 du

S = 4u^3/3 - 8u^5/5 + C

= 4sin^3(x)/3 - 8sin^5(x)/5

but, while other purported solutions may look different, given trig substitutions

they may be equivalent, surprisingly. What looks different may be the same. in the trig world

work a problem differently, and do it correctly, the answer is the same, even if often they

appear different.

Tricky question.

worse than Dr. Nye vs. Dr. Proton on the Big Bang Show

even online integral calculators come up with seemingly different answers.