Trig function Q (1 Viewer)

[miss-smexy]

Hey guys,

can someone please help me with this question:

The diagram shows part of the graph of the function y = tan2x. The shaded region is bounded by the curve, the x-axis, and the line x=pi/6.

The region is rotated about the x-axis to form a solid.

(i) show that the volume of the solid is given by

V = pi (integral between 0 and pi/6) (sec^2 2x-1) dx.

The thing is... how do you integrate:

V = pi (integral between 0 and pi/6) (tan2x)^2 dx

 

[Forbidden.]

You remember the trig identity sin²x + cos²x = 1 ?

Divide all three by cos²x and you get:
tan²x + 1 = sec²x

Make tan²x the subject:
tan²x = sec²x - 1


But tan²2x IS sec²2x - 1
If you find the indefinite integral of sec²2x - 1 you should get (tan 2x)/2 + x + C

Beware of these type of questions in your assessments on calculus involving trigonometric functions, know your trigonometric identities (3 of them).

 

[miss-smexy]

You remember the trig identity sin²x + cos²x = 1 ?

Divide all three by cos²x and you get:
tan²x + 1 = sec²x

Make tan²x the subject:
tan²x = sec²x - 1

Beware of these type of questions in your assessments on calculus involving trigonometric functions, know your trigonometric identities (3 of them).

thanks so much! =)

 

[miss-smexy]

Oh wait, if you solve this question, what answer do you get?
because I got

pi root 3 - ( pi^2 / 6 )


but the answers say:

( pi root 3 / 2 ) - ( pi^2 / 6 )

I did this twice ... so either I am wrong or the answer booklet is? Can someone help!

 

[coblin]

i got 67

 

[watatank]

But tan²2x IS sec²2x - 1
If you find the indefinite integral of sec²2x - 1 you should get

(tan 2x)/2 - x + C

so knowin this, from there, you go

pi * [ (tan 2x)/2 - x ] limits pi/6, 0

= pi * [ ( tan(pi/3) - pi/6 ) - tan0 -0 ]

= pi * [ (sqrt3/2) - (pi/6) ]

= [ pi *( sqrt3/2) - (pi^2/6) ]

 

[miss-smexy]

Ohhh yes I see! thanks! =)

 

[watatank]

Oh wait, if you solve this question, what answer do you get?
because I got

pi root 3 - ( pi^2 / 6 )


but the answers say:

( pi root 3 / 2 ) - ( pi^2 / 6 )

I did this twice ... so either I am wrong or the answer booklet is? Can someone help!

sorry the answer booklet is not wrong. when you do the integral of (sec x)^2 the integral is tan x, but since its (sec2x)^2 you have to divide by two. tip: use differentiation to check back to make sure your integral is correct (for this one you would have to know the chain rule to make sure youre doing it right).

 

[Forbidden.]

sorry the answer booklet is not wrong. when you do the integral of (sec x)^2 the integral is tan x, but since its (sec2x)^2 you have to divide by two. tip: use differentiation to check back to make sure your integral is correct (for this one you would have to know the chain rule to make sure youre doing it right).

I agree with the solution .. I got π[(√3/2) - (π/6)] units³ which is also equal to (π√3/2 - π²/6) units³

mistakes do happen anyway

 

[miss-smexy]

Thanks watatank and f3nr15!
I finally got the answer!