sin‌4(2x)cos6(2x)‌dx Let u=2x, then du=2‌dx, so dx=‌
1
2
du When x=−2π,u=2(−2π)=−4π when x=2π,u=2(2π)=4π I‌=
2Ï€
∫
−2π
sin‌4(2x)cos6(2x)‌dx ‌=‌
1
2
‌
4Ï€
∫
−4π
sin‌4(u)cos6(u)du Since, sin‌(u) and cos(u) are periodic with a period of 2π, So, sin‌4(u) and cos6(u) are also periodic with a period of 2π ‌=‌
1
2
⋅2‌
2Ï€
∫
−2π
sin‌4(u)cos6(u)du ‌=
2Ï€
∫
−2π
sin‌4(u)cos6(u)du ‌=‌
1
2
⋅4‌
2Ï€
∫
0
sin‌4(u)cos6(u)du ‌=2‌
2Ï€
∫
0
sin‌4(u)cos6(u)du Since, the function, f(u)=sin‌4(u)cos6(u) is an even function. ∴‌2‌
2Ï€
∫
0
sin‌4(u)cos6(u)du ‌=2⋅2‌
∫
0
πsin‌4(u)cos6(u)du ‌=4‌
∫
0
πsin‌4(u)cos6(u)du ‌=4⋅2‌
π∕2
∫
0
sin‌4(u)cos6(u)du ‌=8‌
π∕2
∫
0
sin‌4(u)cos6(u)du We know that,
π∕2
∫
0
sin‌m(u)cosn(u)du =‌
(m−1)(m−3)...(1‌ or ‌2)(n−1)(n−3)...(1‌ or ‌2)
(m+n)(m+n−2)...(1‌ or ‌2)
×α where, α=‌
Ï€
2
if both m and n are even, and α=1 otherness. Here, m=4,n=6 both are even. So,