Question

∫▒(sin⁡θ + sin⁡θ tan⁡〖^2 〗 )/sec⁡〖^2 θ〗 dθ
intergrate ...help me please!

Answer 1

Rewrite it properly I can't read it

Answer 2

I=∫▒(sin⁡θ+sin⁡θ (tan⁡θ )^2)/(sec⁡θ )^2 dθ
=∫▒[sec⁡θ tan⁡θ+(sin⁡θ )^3 ] dθ
(sin⁡3θ=3 sin⁡θ-4(sin⁡θ )^3+c
Or~4(sin⁡θ )^3=3 sin⁡θ-sin⁡3θ)
=sec⁡〖θ-3 cos⁡θ 〗+cos⁡3θ/3+c

Answer 3

correction 1/4 outside the bracket after sec theta.

Answer 4

i have another question.. need your help..

Answer 5

by the way, my answer for the integration question is -cos theta..... is it the correct answer...

Answer 6

Let f(x)=sin2x
f(x+h)=sin2(x+h)=sin(2x+2h)=sin2xcos2h+cos2xsin2h
f(x+h)-f(x)=sin2xcos2h-sin2x+cos2xsin2h
=sin2x(cos2h-1)+cos2xsin2h
=sin2x(-2〖sin〗^2h)+cos2xsin2h
=-2sin2x〖sin〗^2h+cos2x sin2h
Divide by h
(f(x+h)-f(x))/h=(-2sin2x〖sin〗^2 h)/h+cos2x sin2⁡h/2h*2
Taking limits as h approaches 0
lim┬(h=0)⁡〖(f(x+h)-f(x))/h=-2sin2x lim┬(h=0)⁡〖(〖sin〗^2 h)/h^2 〗 〗*h+2 cos⁡2x lim┬(h=0)⁡〖sin⁡2h/2h〗
d/dx {f(x)}=-2sin2x*1*0+2 cos⁡2x*1=2cos⁡2x
d/dx sin⁡2x=2 cos⁡2x

Answer 7

can you rewrite back the answer from *divide by h