(1.) ∫sin³ x cos² x dx here is the problem

= ∫(cos² x)(1 - cos² x)(sin x)dx factor like this

= ∫cos² x sin x dx - ∫cos⁴ x sin x dx multiply thru

= (-1/3)cos³ x + (1/5)cos⁵ x + C integrate

(2.) ∫ sin³ x sec x dx here is the problem

= ∫ tan x sec² x dx use this trig id

= (1/2)tan² x + C integrate

(3.) ∫ cos³ x sin^1/2 x dx here is the problem

= ∫(1 - sin² x)(cos x)(sin^1/2 x)dx factor

= ∫sin^1/2 x cos x dx - ∫sin^5/2 x cos x dx multiply thru

= (2/3)sin^3/2 x - (2/7)sin^7/2 x + C integrate

(4.) ∫ cos³ x csc^1/2 x dx here is the problem

= ∫ cos³ x sin^-1/2 x dx reciprocal identity

= ∫(1 - sin² x)(cos x)(sin^-1/2 x)dx trig id, factor

= ∫sin^-1/2 x cos x - ∫sin^3/2 x cos x dx multiply thru

= 2 sin^1/2 x - (2/5)sin^5/2 x + C integrate

(5.) ∫ sin⁵ x cos² x dx here is the problem

= ∫(1 - cos² x)² * (cos² x)(sin x) dx trig id, factor

= ∫(1 - 2 cos² x + cos⁴ x)(cos² x)(sin x)dx square the binomial

= ∫cos² x sin x dx - 2∫cos⁴ x sin x + ∫cos⁶ x sin x dx

[multiply thru parentheses]

= (-1/3)cos³ x + (2/5)cos⁵ x - (1/7)cos⁷ x + C integrate

(6.) ∫ sin⁵ x sec² x dx here is the problem

= ∫(sin⁵ x)(cos^-2 x)dx reciprocal identity

= ∫(1 - cos² x)²(sin x)(cos^-2 x)dx trig id

= ∫(1 - 2cos² x + cos⁴ x)(sin x)(cos^-2 x)dx

[square the binomial]

= ∫cos^-2 x sin x dx - 2∫sin x dx + ∫cos² x sin x dx

[multiply thru parentheses]

= -(cos x)^-1 + 2 cos x - (1/3)cos³ x + C integrate

= - sec x + 2 cos x - (1/3)cos³ x + C reciprocal id

(7.) ∫ sin² x cos⁵ x dx here is the problem

= ∫(sin² x)(1 - sin² x)²(cos x)dx factor, trig id

= ∫(sin² x)(1 - 2sin² x + sin⁴ x)(cos x)dx

[square the binomial]

= ∫sin² x cos x dx - 2∫sin⁴ x cos x dx + ∫sin⁶ x cos x dx

[multiply thru parentheses]

= (1/3)sin³ x - (2/5)sin⁵ x + (1/7)sin⁷ x + C integrate

(8.) ∫ sin² x tan x dx here is the problem

= ∫ sin³ x sec x dx trig id

= ∫(1 - cos² x)(sin x)(cos x)^-1 dx trig id's

= ∫(cos x)^-1 sin x dx - ∫cos x sin x dx multiply thru

= - ln cos x - (1/2)sin² x + C integrate

(9.) ∫ cos² x dx here is the problem

= (1/2)∫(1 + cos 2x)dx half angle id for cos

= (1/2)∫ dx + (1/2)∫cos 2x dx multiply thru parentheses

= (1/2)x + (1/4)sin 2x + C integrate

(10.) ∫ sin⁴ x dx here is the problem

= (1/4)∫(1 - cos 2x)² dx half angle id for sine

= (1/4)∫(1 - 2 cos 2x + cos² 2x) dx square the binomial

= (1/4)∫dx - (1/2)∫cos 2x dx + (1/4)(1/2)∫(1 + cos 4x)dx

[multiply thru parentheses][half angle id for cos]

= (1/4)∫dx - (1/2)∫cos 2x dx + (1/8)∫dx + (1/8)∫cos 4x dx

[multiply thru parentheses]

= (1/4)x - (1/2)(1/2)sin 2x + (1/8)x + (1/8)(1/4)sin 4x + C

= (3/8)x - (1/4)sin 2x + (1/32)sin 4x + C multiply

[and combine like terms]