Compute the following integrals using the guidelines for integrating powers of trigonometric functions. Use a CAS to check the solutions. (Note: Some of the problems may be done using techniques of integration learned previously.) 79. \( \int \sin ^{3} x d x \) 80. \( \int \cos ^{3} x d x \) 81. \( \int \sin x \cos x d x \) 82. \( \int \cos ^{5} x d x \) 83. \( \int \sin ^{5} x \cos ^{2} x d x \) 84. \( \int \sin ^{3} x \cos ^{3} x d x \)

Below are one set of acceptable solutions. (Any answer equivalent to the ones below is correct.) In our work we use the standard techniques for integrating odd powers of sine or cosine and, in the last case, a double‐angle substitution. ───────────────────────────── 79. Integrate ∫ sin³x dx Write sin³x as sin²x·sinx and use the Pythagorean identity:   sin³x = (1 – cos²x) sinx. Then let   u = cos x  ⇒ du = – sin x dx  or sin x dx = –du. Thus,   ∫ sin³x dx = ∫ (1 – cos²x) sin x dx         = –∫ (1 – u²) du         = –(u – u³/3) + C = –cos x + (cos³x)/3 + C. ───────────────────────────── 80. Integrate ∫ cos³x dx Write cos³x as cos²x·cosx and use the identity cos²x = 1 – sin²x:   cos³x = (1 – sin²x) cosx. Let   u = sin x  ⇒ du = cos x dx. Then,   ∫ cos³x dx = ∫ (1 – u²) du = u – u³/3 + C = sin x – (sin³x)/3 + C. ───────────────────────────── 81. Integrate ∫ sin x cos x dx A common substitution is to let u = sin x so that du = cos x dx; alternatively, u = cos x works equally well. For example, let u = sin x:   du = cos x dx  ⇒ cos x dx = du. Thus,   ∫ sin x cos x dx = ∫ u du = u²/2 + C = (sin²x)/2 + C. ───────────────────────────── 82. Integrate ∫ cos⁵x dx Write cos⁵x as cos⁴x · cosx and express cos⁴x in terms of sin x:   cos⁴x = (cos²x)² = (1 – sin²x)². Let   u = sin x  ⇒ du = cos x dx. Then,   ∫ cos⁵x dx = ∫ (1 – u²)² du. Expanding,   (1 – u²)² = 1 – 2u² + u⁴. So,   ∫ cos⁵x dx = ∫ (1 – 2u² + u⁴) du        = u – (2/3)u³ + (1/5)u⁵ + C = sin x – (2 sin³x)/3 + (sin⁵x)/5 + C. ───────────────────────────── 83. Integrate ∫ sin⁵x cos²x dx Since the power of sinx is odd, write sin⁵x = sin⁴x · sin x and use   sin⁴x = (sin²x)² = (1 – cos²x)². Thus,   ∫ sin⁵x cos²x dx = ∫ (1 – cos²x)² sin x · cos²x dx. Now let   u = cos x  ⇒ du = – sin x dx  or sin x dx = –du. Then,   ∫ sin⁵x cos²x dx = –∫ (1 – u²)² u² du. Expand (1 – u²)²:   (1 – u²)² = 1 – 2u² + u⁴. Thus the integrand becomes:   u² – 2u⁴ + u⁶. So,   ∫ sin⁵x cos²x dx = –∫ (u² – 2u⁴ + u⁶) du        = –[ u³/3 – 2u⁵/5 + u⁷/7] + C        = –(cos³x)/3 + (2 cos⁵x)/5 – (cos⁷x)/7 + C. ───────────────────────────── 84. Integrate ∫ sin³x cos³x dx A very neat method is to use a double‐angle substitution. Notice that:   sin³x cos³x = [sin x cos x]·[sin²x cos²x]. But   sin x cos x = (1/2) sin(2x)  and  sin²x cos²x = (1/4) sin²(2x). Thus,   sin³x cos³x = (1/2)(1/4) sin(2x) sin²(2x) = (1/8) sin³(2x). Hence the integral becomes:   ∫ sin³x cos³x dx = (1/8)∫ sin³(2x) dx. Now substitute t = 2x, so that dt = 2 dx or dx = dt/2:   = (1/8) ∫ sin³t · (dt/2) = (1/16) ∫ sin³t dt. We now integrate sin³t dt using the method from (79):   Write sin³t = (1 – cos²t) sin t, and let u = cos t so that du = – sin t dt. Then,   ∫ sin³t dt = –∫ (1 – u²) du = –(u – u³/3) + C = –cos t + (cos³t)/3 + C. Returning to x (with t = 2x):   ∫ sin³x cos³x dx = (1/16)[ –cos(2x) + (cos³(2x))/3] + C         = –[cos(2x)]/16 + [cos³(2x)]/48 + C. ───────────────────────────── Final Answers 79. ∫ sin³x dx = –cos x + (cos³x)/3 + C 80. ∫ cos³x dx = sin x – (sin³x)/3 + C 81. ∫ sin x cos x dx = (sin²x)/2 + C 82. ∫ cos⁵x dx = sin x – (2 sin³x)/3 + (sin⁵x)/5 + C 83. ∫ sin⁵x cos²x dx = –(cos³x)/3 + (2 cos⁵x)/5 – (cos⁷x)/7 + C 84. ∫ sin³x cos³x dx = –cos(2x)/16 + cos³(2x)/48 + C Each of these answers can be verified using a computer‐algebra system.