9:20 AM
Should the newer question be closed as an exact duplicate? It seems to be more about checking the OP's approach thatn about various methods of computing that particular integral.
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We have: $\int \cos^n x\ dx = \frac{1}{n} \cos^{n-1} x \sin x + \frac{n-1}{n}\int \cos^{n-2} x\ dx.$ Find $\int \cos^4x\ dx$ by using the formula twice What I have so far is: $\int \cos^4 x\ dx = \frac{1}{4} \cos^{3} x \sin x + \frac{3}{4}\int \cos^{2} x\ dx$ Now we use the formula for $\int... 1 I want to solve this integral :$$\int \cos^4(x)dx$$ And think about doing the following thing:$\int (1-\sin^2(x))^2dx \rightarrow \int (1-2\sin^2(x)+\sin^4(x)dx$but I think I just complicated it. Any suggestions? Thanks! 9 hours later… 6:22 PM Several candidates to be closed as duplicates of this: 8 (Fitzpatrick Advanced Calculus 2e, Sec. 2.4 #12) For$c \gt 0$, consider the quadratic equation$x^2 - x - c = 0, x > 0$. Define the sequence$\{x_n\}$recursively by fixing$|x_1| \lt c$and then, if$n$is an index for which$x_n$has been defined, defining $$x_{n+1} = \sqrt{c+x_n}$$ Prove ... 0 Let the sequence$\{a_n\}$be defined as$a_1 = \sqrt 2$and$a_{n+1} = \sqrt {2+a_n}$. Show that$a_n \le$2 for all$n$,$a_n$is monotone increasing, and find the limit of$a_n$. I've been working on this problem all night and every approach I've tried has ended in failure. I keep t... Not sure about this (I am not sure whether answers can be generalized to arbitrary$c$). 8 Problem The sequence$(a_n)_{n=1}^\infty$is given by recurrence relation:$a_1=\sqrt2$,$a_{n+1}=\sqrt{2+a_n}$. Evaluate the limit$\lim_{n\to\infty} a_n$. Solution Show that the sequence$(a_n)_{n=1}^\infty$is monotonic. The statement $$V(n): a_n < a_{n+1}$$ holds for$n = 1$, that is ... 2 Given a sequence$\{a_n\}$, with$n\geq1$, where $$a_{1}=4,$$ and $$a_{n+1}=\sqrt{a_{n} +20}.$$ Prove via induction that, for all$n \geq 1$, $$a_{n+1}>a_{n}.$$ Apparent Convergence The sequence appears to be increasing, and possibly bounded at 5. How may I show convergence, ... 3 Computing the first few terms $$a_1=1, a_2=\sqrt{3}=1.732....,a_3=1.9318....,a_4=1.9828...$$ I feel that$(a_n)_{n\in \mathbb{N}}$is bounded above by 2, although I have no logical reasoning for this. Since,$(a_n)_{n\in \mathbb{N}}\$ is monotone increasing sequence, it must converge by monotone ...