integrate sin⁷x.cos⁵x
Answers
Answer ♡
Given
Perimeter of Rectangle is 60m
length is 4m more than four times it's breadth
To findi
length and breadth of the Rectangle
Slove
Let us assume the breadth of Rectangle be 'x'
according to the question
length would be => 4m+4x
Perimeter is 60m
Perimeter of Rectangle=2(length+Breadth)
=>60= 2( 4+4x +x)
=>60=2(4+5x)
=>2(4+5x)=60
=>4+5x=30
=>5x= 30-4
=>5x= 26
=>x= 26÷ 5
=>x=5.2
Now , breadth is 5.2m
length = 4+4x=4+4×5.2
length= 4+20.8=24.8m
Check
Perimeter of Rectangle=2( 24.8+5.2)
Perimeter of Rectangle=2(30)
Perimeter of Rectangle= 60m
Extra information=>
Perimeter is the total distance occupy by a solid 2D figure around its edge.
Area of Rectangle= length × breadth
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hope that u r helpful
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Answer:
Your objective is to express these in terms of sine/cosine with terms of power 1.
Note that:
cos^5(x)
= cos^2(x) * cos^2(x) * cos(x)
= [1 + cos(2x)]/2 * [1 + cos(2x)]/2 * cos(x)
= [cos^2(2x) + 2cos(2x) + 1]/4 * cos(x)
= [1/2 + (1/2)cos(4x) + 2cos(2x) + 1]/4 * cos(x)
= [(1/2)cos(4x)cos(x) + 2cos(2x)cos(x) + (3/2)cos(x)]/4
= (1/8)cos(4x)cos(x) + (1/2)cos(2x)cos(x) + (3/8)cos(x)
= (1/16)[cos(4x + x) + cos(4x - x)] + (1/4)[cos(2x + x) + cos(2x - x)] + (3/8)cos(x)
= (1/16)cos(5x) + (5/16)cos(3x) + (5/8)cos(x)
Integrating this yields (1/80)sin(5x) + (5/48)sin(3x) + (5/8)sin(x) + C.
For sin^7(x), we do a similar approach to yield:
sin^7(x)
= sin^2(x) * sin^2(x) * sin^2(x) * sin(x)
= [1 - cos(2x)]/2 * [1 - cos(2x)]/2 * [1 - cos(2x)]/2 * sin(x)
= [-cos^3(2x) + 3cos^2(2x) - 3cos(2x) + 1]/8 * sin(x)
= [(-3/4)cos(2x) - (1/4)cos(6x) + 3/2 + (3/2)cos(4x) - 3cos(2x) + 1]/8 * sin(x)
= (-15/32)cos(2x)sin(x) - (1/32)cos(6x)sin(x) + (3/16)cos(4x)sin(x) + (5/8)sin(x)
= (35/64)sin(x) - (21/64)sin(3x) + (7/64)sin(5x) - (1/64)sin(7x).
Finally, integrating this yields (-35/64)cos(x) + (7/64)cos(3x) - (7/320)cos(5x) + (1/448)cos(7x) + C.
I hope this helps!