Assalamu'alaikum , Good Morning all..!
I have a problem, can you check or correct this? maybe I had do something wrong
∫( sin^4 x + cos^4 x + 2 sin^2 x . cos^2 x ) dx = .....
A. 1/5 cos^5 x - 1/5 sin^5 x + c
B. 1/5 cos^5 x + 1/5 sin^5 x + c
C. 1/5 cos^5 x - 1/5 sin^5 x + 2/3 sin^3 x . cos^3 x + c
D. x sin x + c
E. x + c
My Answer:
∫( sin^4 x + cos^4 x + 2 sin^2 x . cos^2 x ) dx =
= ∫ sin^4 x dx + ∫cos^4 x dx + ∫ 2 sin^2 x . cos^2 x dx =
First, I solve ∫ sin^4 x dx = ∫(sin^2 x)^2 dx
= ∫ [1/2 (1 - cos 2x)]^2 dx
= ∫ 1/4 (1 - 2 cos 2x + cos^2 2x) dx
= ∫ 1/4 - 2/4 cos 2x + 1/4 cos^2 2x dx
= ∫ 1/4 - 1/2 cos 2x + 1/4 cos^2 2x dx
= 1/4x - 1/4 sin 2x + ∫ 1/4 cos^2 2x dx
= 1/4x - 1/4 sin 2x + ∫ 1/4 . 1/2 (1 + cos 2(2x)) dx
= 1/4x - 1/4 sin 2x + ∫ 1/8 (1 + cos 4x) dx
= 1/4x - 1/4 sin 2x + 1/8x + 1/8 . 1/4 sin 4x + c
= 1/4x - 1/4 sin 2x + 1/8x + 1/32 sin 4x + c
∫ sin^4 x dx = 3/8x - 1/4 sin 2x + 1/32 sin 4x + c
Second, I solve ∫ cos^4 dx = ∫(cos^2)^2 dx
= ∫ [1/2 (1 + cos 2x) ]^2 dx
= ∫ 1/4 (1 + 2 cos 2x + cos^2 2x) dx
= ∫ 1/4 + 2/4 cos 2x + 1/4 cos^2 2x dx
= 1/4x + 1/4 sin 2x + ∫ 1/4 cos^2 2x dx
= 1/4x + 1/4 sin 2x + ∫ 1/4 . 1/2 (1 + cos 2(2x)) dx
= 1/4x + 1/4 sin 2x + ∫ 1/8 (1 + cos 4 x) dx
= 1/4x + 1/4 sin 2x + 1/8x + 1/8 . 1/4 sin 4x + c
= 1/4x + 1/4 sin 2x + 1/8x + 1/32 sin 4x + c
∫ cos^4 dx = 3/8x + 1/4 sin 2x + 1/32 sin 4x + c
Third, I solve ∫ 2 sin^2 x . cos^2 x dx = ∫ [sin (x + x) + sin (x - x)]^2 dx
= ∫ [sin 2x + 0]^2 dx
= ∫ sin^2 2x dx
= ∫ 1/2 (1 - cos 2(2x)) dx
= ∫ 1/2 (1 - cos 4x) dx
= 1/2x - 1/2 . 1/4 sin 4x + c
∫ 2 sin^2 x . cos^2 x dx = 1/2x - 1/8 sin 4x + c
After I solve each integrals, I combine all be one, ....
∫( sin^4 x + cos^4 x + 2 sin^2 x . cos^2 x ) dx =
= ∫ sin^4 x dx + ∫cos^4 x dx + ∫ 2 sin^2 x . cos^2 x dx =
= [(3/8x - 1/4 sin 2x + 1/32 sin 4x + c) + (3/8x + 1/4 sin 2x + 1/32 sin 4x + c) +
(1/2x - 1/8 sin 4x + c)]
= [3/8x - 1/4 sin 2x + 1/32 sin 4x + 3/8x + 1/4 sin 2x + 1/32 sin 4x +
1/2x - 1/8 sin 4x] + c
= [3/8x + 3/8x + 1/2x - 1/4 sin 2x + 1/4 sin 2x + 1/32 sin 4x + 1/32 sin 4x - 1/8 sin 4x] + c
= [10/8x + 0 - 2/32 sin 4x] +c = 10/8x - 2/32 sin 4x + c
So, ∫( sin^4 x + cos^4 x + 2 sin^2 x . cos^2 x ) dx = 10/8x - 2/32 sin 4x + c
look, I didn't find the answer , how? help me please...
I have a problem, can you check or correct this? maybe I had do something wrong
∫( sin^4 x + cos^4 x + 2 sin^2 x . cos^2 x ) dx = .....
A. 1/5 cos^5 x - 1/5 sin^5 x + c
B. 1/5 cos^5 x + 1/5 sin^5 x + c
C. 1/5 cos^5 x - 1/5 sin^5 x + 2/3 sin^3 x . cos^3 x + c
D. x sin x + c
E. x + c
My Answer:
∫( sin^4 x + cos^4 x + 2 sin^2 x . cos^2 x ) dx =
= ∫ sin^4 x dx + ∫cos^4 x dx + ∫ 2 sin^2 x . cos^2 x dx =
First, I solve ∫ sin^4 x dx = ∫(sin^2 x)^2 dx
= ∫ [1/2 (1 - cos 2x)]^2 dx
= ∫ 1/4 (1 - 2 cos 2x + cos^2 2x) dx
= ∫ 1/4 - 2/4 cos 2x + 1/4 cos^2 2x dx
= ∫ 1/4 - 1/2 cos 2x + 1/4 cos^2 2x dx
= 1/4x - 1/4 sin 2x + ∫ 1/4 cos^2 2x dx
= 1/4x - 1/4 sin 2x + ∫ 1/4 . 1/2 (1 + cos 2(2x)) dx
= 1/4x - 1/4 sin 2x + ∫ 1/8 (1 + cos 4x) dx
= 1/4x - 1/4 sin 2x + 1/8x + 1/8 . 1/4 sin 4x + c
= 1/4x - 1/4 sin 2x + 1/8x + 1/32 sin 4x + c
∫ sin^4 x dx = 3/8x - 1/4 sin 2x + 1/32 sin 4x + c
Second, I solve ∫ cos^4 dx = ∫(cos^2)^2 dx
= ∫ [1/2 (1 + cos 2x) ]^2 dx
= ∫ 1/4 (1 + 2 cos 2x + cos^2 2x) dx
= ∫ 1/4 + 2/4 cos 2x + 1/4 cos^2 2x dx
= 1/4x + 1/4 sin 2x + ∫ 1/4 cos^2 2x dx
= 1/4x + 1/4 sin 2x + ∫ 1/4 . 1/2 (1 + cos 2(2x)) dx
= 1/4x + 1/4 sin 2x + ∫ 1/8 (1 + cos 4 x) dx
= 1/4x + 1/4 sin 2x + 1/8x + 1/8 . 1/4 sin 4x + c
= 1/4x + 1/4 sin 2x + 1/8x + 1/32 sin 4x + c
∫ cos^4 dx = 3/8x + 1/4 sin 2x + 1/32 sin 4x + c
Third, I solve ∫ 2 sin^2 x . cos^2 x dx = ∫ [sin (x + x) + sin (x - x)]^2 dx
= ∫ [sin 2x + 0]^2 dx
= ∫ sin^2 2x dx
= ∫ 1/2 (1 - cos 2(2x)) dx
= ∫ 1/2 (1 - cos 4x) dx
= 1/2x - 1/2 . 1/4 sin 4x + c
∫ 2 sin^2 x . cos^2 x dx = 1/2x - 1/8 sin 4x + c
After I solve each integrals, I combine all be one, ....
∫( sin^4 x + cos^4 x + 2 sin^2 x . cos^2 x ) dx =
= ∫ sin^4 x dx + ∫cos^4 x dx + ∫ 2 sin^2 x . cos^2 x dx =
= [(3/8x - 1/4 sin 2x + 1/32 sin 4x + c) + (3/8x + 1/4 sin 2x + 1/32 sin 4x + c) +
(1/2x - 1/8 sin 4x + c)]
= [3/8x - 1/4 sin 2x + 1/32 sin 4x + 3/8x + 1/4 sin 2x + 1/32 sin 4x +
1/2x - 1/8 sin 4x] + c
= [3/8x + 3/8x + 1/2x - 1/4 sin 2x + 1/4 sin 2x + 1/32 sin 4x + 1/32 sin 4x - 1/8 sin 4x] + c
= [10/8x + 0 - 2/32 sin 4x] +c = 10/8x - 2/32 sin 4x + c
So, ∫( sin^4 x + cos^4 x + 2 sin^2 x . cos^2 x ) dx = 10/8x - 2/32 sin 4x + c
look, I didn't find the answer , how? help me please...