Can you check my work? I didnt find right answer, please help me, thanks you.

latifah12

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Aug 7, 2013
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2
Assalamu'alaikum :), Good Morning all..!

I have a problem, can you check or correct this? maybe I had do something wrong :confused:

∫( 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...
 
Assalamu'alaikum :), Good Morning all..!

I have a problem, can you check or correct this? maybe I had do something wrong :confused:

∫( 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...

\(\displaystyle \displaystyle \int \left [cos^4(x) + sin^4(x) + 2*cos^2(x)*sin^2(x)\right ]dx\)


\(\displaystyle = \displaystyle \int \left [cos^2(x) + sin^2(x)\right ]^2dx\)

do you see a much shorter way now.....
 
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