Question

Using substitution cosx=u
integrate sin3x.cos4x
and show it is equal to
17cos7x−15cos5x+c

Answer 1

Sorry the 17cos7x−15cos5x+c is wrong it's meant to be 1/7cos^7 x -1/5cos^5 x

Answer 2

lol. that changes everything :)

let u = cos x ---> du = -sinx dx

Answer 3

remember that sin^2(x) = 1 - cos^2(x). you must use this identity when doing trig integration.

so we have (1 - cos^2(x))*cos^4(x)*sinx dx = (1 - u^2)*u^4 du

Answer 4

Also the top bit is sin^3 x cos^4 x, not sure why it didn't go to a power when writing the equation

Answer 5

what i said times - 1 since du = - sinx

Answer 6

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

Answer 7

should i keep going?

Answer 8

Yes I understand so far thanks

Answer 9

not sure if if that means i should or not :P

Answer 10

yes keep going please :p

Answer 11

= integral of u^6 - u^4
= (1/7)u^7 - (1/5)u^5 + C

plug back u = cos x

= (1/7)cos^7(x) - (1/5)cos^5(x) + C

Answer 12

better we use the identity :
sinAcosB = 1/2 (sin(A+B)+sin(A-B))

so, sin3x cos4x = 1/2 (sin(3x+4x) + sin(3x-4x) = 1/2 (sin7x - sinx)
now, see ur integral becomes