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INTEGRALS INVOLVING COMPOUND ANGLE IDENTITIES

Problem 1 :

𝜋601 - cos 6x dx

Solution :

Given, 𝜋601 - cos 6x dx= 𝜋602 sin23x dx= 2 𝜋60 sin 3x dx= 2 -cos 3x3𝜋60= 2 13 -cos 3𝜋6 + 13 cos 3(0) = 2 13 -cos 𝜋2 + 13 cos 0 = 2 0 + 13= 23

Problem 2 :

sin 2a cos 8a da

Solution :

sin 2a cos 8a da=sin 2a cos 8a dasin A cos B = 12[sin(A+B) + sin (A -B)]= 12[sin(2a +8a) + sin (2a -8a)]= 12[sin 10a+ sin (-6a)]=12 [sin 10a- sin (6a)]= 12-cos10a10+cos 6a 6+C= -cos10a20+cos 6a 12+C

Problem 3 :

cos b cos 11 b db

Solution :

Given, cos b cos 11 b dbcos A cos B = 12[cos (A + B) - cos (A - B)]= 12(cos (b + 11b) - cos (b - 11b))db= 12 (cos 12b - cos (-10b)) db= 12 cos 12b - cos (-10b) db= 12 cos 12b db -cos (-10b) db = 12 112(sin 12b) + 110(sin 10b)= 124sin 12b + 120sin 10b= 124sin 12b + 120sin 10b + C

Problem 4 :

sin 6𝜃 sin 14𝜃 d𝜃

Solution :

Given, sin 6𝜃 sin 14𝜃 d𝜃sin A sin B = 12[cos (A - B) + cos (A + B)]= 12(cos (6𝜃 - 14𝜃) + cos (6𝜃 + 14𝜃))d𝜃= 12 cos (-8𝜃) + cos (20𝜃) d𝜃= 12 cos (-8𝜃) + cos 20𝜃 d𝜃= 12 cos (-8𝜃) d𝜃 +cos 20𝜃 d𝜃 = 12 18sin(-8𝜃) + 120sin 20𝜃= -116sin 8𝜃 + 140sin 20𝜃= -116sin 8𝜃 + 140sin 20𝜃+ C

Problem 5 :

sin 7x cos 3x dx

Solution :

Given, sin 7x cos 3x dxsin A cos B = 12[sin (A + B) + sin (A - B)]= 12(sin (7x + 3x) + sin (7x - 3x))dx= 12 sin (10x) + sin (4x) dx= 12 sin 10x + sin 4x dx= 12 sin 10x dx +sin 4x dx = 12 110(-cos 10x) + 14(-cos 4x)= 120(-cos10x) - 18cos 4x= -120cos 10x - 18cos 4x + C

Problem 6 :

sin 10x sin 4x dx

Solution :

Given, sin 10x sin 4x dxsin A sin B = 12[cos (A - B) + cos (A + B)]= 12(cos (10x - 4x) + cos (10x + 4x))dx= 12 cos (6x) + cos (14x)) dx= 12 cos 6x + cos 14x dx= 12 cos 6x dx +cos 14x dx = 12 16sin 6x + 114sin 14x]= 112sin 6x - 128sin 14x= 112sin 6x - 128sin 14x + C
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