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First assume following relations:

1) sin^2X = 1 - cos^2X and cos^2X = 1 - sin^2X

2) sin^2X = ( 1 - cos2X ) / 2

3) cos^2X = ( 1 + cos2X ) / 2

Now :

sin^5X = sin^X * ( 1 - cos^2X )^2 = sinX * ( 1 - 2cos^2X = cos^4X ) = sinX - 2sinXcos^2X + sinXcos^4X

Integrating this seems simple:

-cosX + 2/3cos^3X - 1/5cos^5x

Integrating cos^5X is just the same, you will have this expression:

cos^5X = cosX - 2cosXsin^2X + cosXsin^4X

for sin^6X and cos^6X you can use 2nd and 3rd formulas to redusing it into espressions with lower degrees :

sin^6X = ( sin^2X )^3

= ( ( 1 - cos2X )/2 )^3

= 1/8( 1 - cos2X)^3

= 1/8( 1 - 3cos2X + 3cos^2(2X) - cos^3(2x ) ) : (1)

Now notice that :

I) cos^2t = ( 1 + cos2t ) / 2

>> Integrate : 1/2( t + 1/2sin2t )

II) cos^3t = cos(t)( 1 - sin^2t )

= cos(t) - cos(t)*sin^2(t)

>> Integrate : sin(t) - 1/3sin^3(t)

Now that you can Integrate cos^2(t) and cos^3(t) you can also integrate expression (1)

this is also the same for cos^6X, using formula (3) you will have :

cos^6X = 1/8( 1 + 3cos2X + 3cos^2(2X) + cos^3(2x ) )

Calculating the integral with limits seems quite easy.

hope it was helpful.

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The trig one: This is rather tedious, but the answer is 45Pi/65536 for the first one (i'm assuming the upper integration limit is Pi/2 and the lower is 0. If the limits are reversed, then the answer is the negative of the above) The other gives 2*x+2*arctan(x),which, evaluated from 6 to 10, gives 8+2*arctan(10)-2*arctan(6) or 8.13096

Sin 6 X

a) let y = sin^5 X with limit 0 to pi/2

integral of y = 5 sin^4 x cos x. . . . substitute the limit

. . . . . = 5 ( sin^4 90 cos 90 - sin^4 0 cos 0)

. . . . . = 5 ( 0 - 0 ) = 0

b) let y = sin^6 X with limit 0 to pi/2

integral of y = 6 sin^5 x cos x. . . . substitute the limit

. . . . . = 6 ( sin^5 90 cos 90 - sin^5 0 cos 0)

. . . . . = 6 ( 0 - 0 ) = 0

c) let y = cos^5 X with limit 0 to pi/2

integral of y = 5 cos^4 x (-sin x ). . . . substitute the limit

. . . . . = 5 ( cos^4 90 (- sin 90 ) - cos^4 0 (- sin 0 ))

. . . . . = 5 ( 0 - 0 ) = 0

d) let y = cos^6 X with limit 0 to pi/2

integral of y = 6 cos^5 x (- sin x ). . . . substitute the limit

. . . . . = 6 ( cos^5 90 (- sin 90 ) - cos^5 0 (- sin 0))

. . . . . = 6 ( 0 - 0 ) = 0

i can only help you in the last bit. ummm im a lil' rusty.

0 to pie/2 is the same as 90 degrees.

so when you get the answers for each individual term, it must be between that interval.

0 - 90 degrees. if it exceeds or preceeds the limit then disregard the term.