Question

1. integral 1 / cos^6(3x + 2) = ?
2. integral cos^2 3x dx = ?
3. integral sin^2 dx = ?

Answer 1

1) int 1/cos^6 (3x+2) dx
= int sec^6 (3x+2) dx
misal :
u = 3x + 2
du = 3 dx ----> dx = 1/3  du
so,
=1/3  int sec^6 (u) du
= 1/3 int (sec^2(u))^2 sec^2(u) du
= 1/3 int (1+tan^2(u))^2 sec^2(u) du
= 1/3 int (1 + 2 tan^2(u) + tan^4(u)) sec^2u) du
split menjadi 3 integral :
= 1/3 int sec^2(u) du + 2/3 int tan^2(u) sec^2(u) du + 1/3 int tan^4(u) sec^2(u) du

jelas, untuk 1/3 int sec^2(u) du = 1/3 tan(u)
untuk yg lainnya gunakan int u-subs,
misal y = tan(u)
dy = sec^2(u) du
shg, diperoleh :
2/3 int tan^2(u)sec^2(u)du + 1/3 int tan^4(u)sec^2(u) du
= 2/3 int y^2 dy + 1/3 y^4 dy
= 2/3 * 1/3 y^3 + 1/3 * 1/5 y^5 + C
= 2/9 y^3 + 1/15 y^5 + c
substitusikan kembali bahwa y = tan(u), maka
= 2/9 tan^3 (u) + 1/15 tan^5 (u)

shg totalnya,
1/3 tan(u) + 2/9 tan^3 (u) + 1/15 tan^5 (u) + C
finally, substitusikan kembali u = (3x + 2)
so,
= 1/3 tan(3x+2) + 2/9 tan^3 (3x+2) + 1/15 tan^5 (3x+2) + C

2) int cos^2 (3x) dx
= int (1 + cos(6x))/2 dx
= int 1/2 dx + 1/2 cos(6x) dx
= 1/2 x + 1/12 sin(6x) + C

3) int sin^2 apa ya .... hehe, i was tired :)