Intergrating Tirgometric Functions (1 Viewer)

[davidbarnes]

Just started to intergrate trig and I am having some trouble with it. I can do the basic questions but are having soem trouble with these harder questions. I can't seem to get the correct answer at all. If someone could show me how to solve the below it woudl be very appreciated thanks.

Differerntiate
1. Sin^2 X
2. 3Cos^3 5X
3. CosXSin^4 X

 

[powerdrive]

Just started to intergrate trig and I am having some trouble with it. I can do the basic questions but are having soem trouble with these harder questions. I can't seem to get the correct answer at all. If someone could show me how to solve the below it woudl be very appreciated thanks.

Differerntiate
1. Sin^2 X
2. 3Cos^3 5X
3. CosXSin^4 X<!-- google_ad_section_end -->

1. d/dx (sin²x)
= d/dx (sinx)²
= 2sinx × cosx
= sin2x

→using: d/dx [f(x)]^n = n × [f(x)]^n-1 × f '(x)

2. d/dx (3cos³5x)
= d/dx [3(cos5x)³]
= 3 × 3cos²5x × -5sin5x
= -45sin5x cos²5x

3. d/dx (cosx sin^4 x)
= d/dx [cosx (sinx)^4]
= uv’ + vu’ ; where u=cosx, v=sin^4 x
= 4sin³x cosx × cosx + -sinx × sin^4 x
= 4sin³x cos²x – sin^5 x

→using: d/dx (uv) = uv' + vu'

 

[davidbarnes]

1. d/dx (sin²x)
= d/dx (sinx)²
= 2sinx × cosx
= sin2x

Thanks for your help Powerdrive. With #1 why would the answer be sin2X and not 2sinXcosX?

 

[davidbarnes]

Got a few more questions if anyone is up to them.

@ = pie

1. Find the gradient of the tangent to the curve y=tan3X at the point where X = @/9

2. Find the equation of the tangent to the curve y=sin(@-X0 at the point (@/6, 1/2), in exact form.

3. Find the equation of the normal to the curve y=3sin2X at the point where X = @/8, in exact form.

4. If y=2sin3X-5cos3X, shwo that d^2y/dx^2 = -9y

 

[powerdrive]

Thanks for your help Powerdrive. With #1 why would the answer be sin2X and not 2sinXcosX?

well see sin2x = 2sinxcosx, you can leave the answer as either of them.

 

[Mark576]

1. dy/dx = 3sec²3x

At x = pi/9; dy/dx = 3[1/cos²(pi/3)] = 12

2. dy/dx = -cos(pi-x), at x = pi/6;

dy/dx = -cos(pi-pi/6) = &radic;3/2

&there4; Equation of tangent at (pi/6 , 1/2) is given by:

y - 1/2 = &radic;3/2 (x - pi/6) <-- you can simplify that further if you wish

3. dy/dx = 6cos2x, at x = pi/8, dy/dx = 6/&radic;2, so for the gradient of the normal, m = -&radic;2/6. I'm sure you can then finish off this question.

4. dy/dx = 6cos3x + 15sin3x

d²y/dx² = -18sin3x + 45 cos3x = -9(2sin3X-5cos3X) = -9y