- #1
tandoorichicken
- 245
- 0
How do you integrate even powers of trig functions, such as
\int\cos^{2}{\theta} \,d\theta
\int\cos^{2}{\theta} \,d\theta
Integrate Even Powers of Trig Functions: Cos²θ
- Thread starter tandoorichicken
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AI Thread Summary
To integrate even powers of trigonometric functions like cos²θ, it is effective to use the identity cos²θ = (1 + cos 2θ)/2. For higher even powers, such as cos⁴θ, the process involves squaring the identity, resulting in cos⁴θ = ((1 + cos 2θ)/2)². Similarly, the identity sin²θ = (1 - cos 2θ)/2 can be applied for sine functions. It is important to remember to double the angle when using these identities. This method simplifies the integration process significantly.
- #2
phy
Just an idea: perhaps you could break the integral into cos(theta)*cos(theta) and then integrate by parts from there 
- #3
relinquished™
- 79
- 0
tandoorichciken, you must use an identity
<br /> <br /> cos^2 \theta = \frac{1 + cos 2\theta}{2}<br /> <br />
When the cosine function is raised to an even integer, say for example 4,
<br /> cos^4 \theta = (cos^2)^2 = (\frac{1 + cos 2\theta}{2})^2<br />
For the sine function, use the identity
<br /> <br /> sin^2 \theta = \frac{1 - cos 2\theta}{2}<br /> <br />
A word of Caution. When doing this method do not forget to double your angle ^_^
<br /> <br /> cos^2 \theta = \frac{1 + cos 2\theta}{2}<br /> <br />
When the cosine function is raised to an even integer, say for example 4,
<br /> cos^4 \theta = (cos^2)^2 = (\frac{1 + cos 2\theta}{2})^2<br />
For the sine function, use the identity
<br /> <br /> sin^2 \theta = \frac{1 - cos 2\theta}{2}<br /> <br />
A word of Caution. When doing this method do not forget to double your angle ^_^
Neglect gravity other than that of water; neglect friction.
F1=ρgsh=1000*10*1*(1+4)=50000 N;
F1=ρgsh=1000*10*0.1*4=4000 N;
F3=50000-4000=46000 N?
TL;DR Summary: I came across this question from a Sri Lankan A-level textbook.
Question - An ice cube with a length of 10 cm is immersed in water at 0 °C. An observer observes the ice cube from the water, and it seems to be 7.75 cm long. If the refractive index of water is 4/3, find the height of the ice cube immersed in the water.
I could not understand how the apparent height of the ice cube in the water depends on the height of the ice cube immersed in the water. Does anyone have an...
Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system
$$M(t) = M_{C} + m(t)$$
$$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$
$$P_i = Mv + u \, dm$$
$$P_f = (M + dm)(v + dv)$$
$$\Delta P = M \, dv + (v - u) \, dm$$
$$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$
$$F = u \frac{dm}{dt} = \rho A u^2$$
from conservation of momentum , the cannon recoils with the same force which it applies.
$$\quad \frac{dm}{dt}...
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