DEFINITE INTEGRAL OF TRIGONOMETRIC FUNCTIONS

Problem 1 :

𝜋0(1 + sin x) dx

Solution :

Given, 𝜋0(1 + sin x) dx=𝜋01 dx + 𝜋0 sin x dx= [x]𝜋0 + [-cos x]𝜋0= 𝜋 + (-cos 𝜋 -(-cos 0))= 𝜋 - cos 𝜋 + cos 0= 𝜋 - (-1) + 1= 𝜋 + 2

Problem 2 :

𝜋40 1 - sin2𝜃cos2𝜃 d𝜃

Solution :

Given, 𝜋40 1 - sin2𝜃cos2𝜃 d𝜃= 𝜋40 cos2 𝜃cos2 𝜃 d𝜃= 𝜋40 1 d𝜃= [𝜃]𝜋40= 𝜋4 - 0= 𝜋4

Problem 3 :

𝜋6-𝜋6 sec2 x dx

Solution :

Given, 𝜋6-𝜋6 sec2 x dx= [tan x]𝜋6-𝜋6= tan 𝜋6 - tan -𝜋6= 33 - -33= 3 + 33= 233= 233 · 3= 23

Problem 4 :

𝜋2𝜋4 2 - csc2 x dx

Solution :

Given, 𝜋2𝜋4 2 - csc2 x dx = 𝜋2𝜋4 2 dx - 𝜋2𝜋4 csc2 x dx = [2x]𝜋2𝜋4 - [-cot x]𝜋2𝜋4 = 2𝜋2 - 2𝜋4 - -cot 𝜋2 + cot 𝜋4= 𝜋 - 𝜋2 - [-0 + 1]= 𝜋 - 𝜋2 -1= 2𝜋 - 𝜋2 - 1= 𝜋2 - 1

Problem 5 :

𝜋3-𝜋3 4 sec 𝜃 tan 𝜃 d𝜃

Solution :

Given, 𝜋3-𝜋3 4 sec 𝜃 tan 𝜃 d𝜃 = 4[sec 𝜃]𝜋3-𝜋3= 4sec 𝜋3 - sec -𝜋3= 4[2 - 2]= 4(0)= 0

Problem 6 :

𝜋2-𝜋2(2t+ cos t) dt

Solution :

Problem 7 :

1-1e𝜃 + sin 𝜃 d𝜃

Solution :

Given, 1-1e𝜃 + sin 𝜃 d𝜃 = 1-1e𝜃 d𝜃 + 1-1sin 𝜃 d𝜃 = e𝜃1-1 - [cos 𝜃]1-1= e1 - e-1 - (cos (1) - cos (-1))= e1 - e-1 - cos (1) + cos (1) = e - e-1

Problem 8 :

2eecos x - 1x dx

Solution :

Given, 2eecos x - 1x dx= cos x - 1x dx=cos x dx - 1x dx= [sin x - In |x|]2ee= [(sin (2e) - In|2e|) -(sin e - In|e|]= sin (2e) - In|2e| - sin e + In|e|= sin 2e - sin e + In |e||2e|= sin 2e - sin e + In 12= sin 2e - sin e - In 2

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