x*dx Using trigonometric substitution technique, | the simplified integral after substitution is: 4 A J16sin*ecosede ® J16sec ede © S= 16sece de tane O J16sec etanede

Q: Given that cos 0 terminates in quadrant I, then what is the 40 41 A and that angle 0 value of sin 0?

A: See the answer below

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A: Since you have asked multiple question, we will solve the first question for you as per our guide…

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A: ∫(tan2u)du then put x=tan(u) then u=tan-1(x) and du=dxx2+1

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A: The value of the given indefinite integral is (1/4)ln(|cot x|) - (1/8)ln(|cot²x + 4|) + C

Q: Evaluate: Je*cos(7e*)cos(4e*)dx. 1 - cos(3e") - -cos(11e*) +C 22 sin(3e") - sin(11e*) + c 22…

A:

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A: To find whether given integral can be solved only by algebraic substituion technique

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Q: Evaluate: Jy?cos2ydy 글an2y-글ycos2y+ 금snzy+ A 1 1 1 żsin2y-ycos2y-sin2y+C В 4 1 © sin2y+ycos2y+sin2y+…

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A: To solve the following integral by substitution

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Q: Find the principle solution of tan(4 – 4x)= 0. %3D

A: find for x tan(4 - 4x) = 0 4 - 4x = tan-(0)              [ take both side tan inverse 4 - 4x = 0…

Q: x*dx , the simplified integral after substitution is: |x² – 4 Using trigonometric substitution…

A: To simplify the above integral we have to use trigonometric technique .

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A: To find the area between the parametric curves

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A: We have to show that f and g are inverse function.

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A: Let us take  A = (−1,2,4), B = (0,4,8), and C(1,1,5).

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A: answer is in next step

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A: Follow the procedure given below.

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A: A detailed solution is given below

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Q: Evaluate: Je*cos(7e*)cos(4e*)dx. ④ 금Sn(3e)- sin(11e")+C sin(11e*) + C 22 A © -cos(ae)- cos(11e) +c…

A: The given integral,   ∫excos7excos4exdx

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A:

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A: Given problem :-

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A: Here we have to find the area of triangle.

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Q: Question 15 If h(x): cosxdx and h =4, then h(0) is equal to A B) None -4 D. -3 E

A: