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Equal sign is correctly used here. Greater than or equal to sign is still mathematically correct but not preferable. If someone claims t>=u=v, it means t>=u and u=v.

Even in more complex examples, it is still advisable to use the cover up method as the first step to reduce the number of unknowns. Example 1 1/[(x+1) (x-1) (x-2)^2] = A/(x+1) + B/(x-1) + C/(x-2) + D/(x-2)^2 By cover up method, A=1/[(-1-1) (-1-2)^2]=-1/18 B=1/[(1+1) (1-2)^2]=1/2 D=1/[(2+1)...

It looks like solving a differential equation by integrating factor.

Long division followed by partial fraction decomposition is the most standard approach for integrating rational functions. Alternatively, some people may also use partial fraction decomposition directly by noting the degree of the quotient. Let x^4/(x^2-3x+2)=Ax^2+Bx+C+D/(x-1)+E/(x-2).

What's the point to create two separate formula when you can simply replace a^2 and -a^2 by a real constant k to get a single formula? Differentiating RHS will give you LHS readily.

Cover up method only works for a linear factor in its highest degree. Consider 1/[(x-1) (x-2)^2 (x-3)^3 (x^2+x+1)] = A/(x-1) + B/(x-2) + C/(x-2)^2 + D/(x-3) + E/(x-3)^2 + F/(x-3)^3 + (Gx+H)/(x^2+x+1). It only works for A, C and F.

Let u=sqrt(e^(2x)-1) x=(1/2)ln(u^2+1) dx=u/(u^2+1) du

Let u=sqrt(x) u^2=x 2u du=dx

\int_{0}^{\pi}\left(3\pi-\cos x-2x\right)\left(\frac{x}{1+\sin x}\right)^{2}dx

There is the word "hence" so you may consider using (ii). Aftet applying (ii), you get a difference of sine in the numerator and you can now apply (i). sin(9x+8x)-sin(9x-8x) If you are familiar with double angle formula, you may recognize sin8x=8 sinx cosx cos2x cos4x.

The proposition is not defined for n=1. You are the required to prove the proposition for n>=2 and the base case is n=2, which has only 1 term.

Does this one require a little bit of hand-holding?:p \int_{\frac{\pi}{10}}^{\frac{3\pi}{20}}\ln\left(\frac{1}{4}+\frac{3\tan x-1}{1-3\tan^{2}x}\right)dx By the way, there are variants that I often find more useful...

Graphing the integral (anti-derivative) doesn't add much difficulty. Suppose you are given f'(x) and you need to graph f(x). When f'(x) is positive, f(x) is increasing (and vice versa). When f'(x) is increasing, f(x) is concave up (and vice versa).

\int_{0}^{\frac{\pi}{8}}\frac{\left(\sin x-\cos x\right)\ln\left(\tan2x\right)}{\left(\sin x+\cos x\right)\sqrt{\cos4x}}dx=\frac{3\pi\ln2}{16}

These are different stronger versions of induction. A. The statement is true for n=1 If it is true for n=1,2......,k, then it is also true for n=k+1. B. The statement is true for n=1,2 If it is true for n=k,k+1, then it is also true for n=k+2.

This is essentially a watered-down version of another question that I had seen before. Prove that 2903^n – 803^n – 464^n + 261^n is divisible by 1897 for any positive integers. This kind of problem can be solved by number theory in a much more elegant way. It is just not worthwhile to restrict...

Feel free to share your attempt. \int_{\frac{1}{4}}^{\frac{1}{2}}\frac{1+2^{8x-1}\cos x^{x}+\ln x+256^{x}+\left(\cos\frac{16^{x}\pi}{3}\right)\ln\sqrt{ex}}{8+4\cos\frac{16^{x}\pi}{3}+4\cos...

a slight variant \int_{0}^{\frac{\pi}{4}}\frac{\left(\sin2x\right)\left(\sin^{-1}\left(\cos^{2}x\right)\right)}{2+\sin x+\sqrt{1+\cos^{2}x}}dx=\sqrt{2}-\sqrt{6}+\frac{\left(-2+11\sqrt{2}+2\sqrt{3}-3\sqrt{6}\right)\pi}{12}+\ln\left(20-12\sqrt{2}+10\sqrt{3}-8\sqrt{6}\right)-\frac{\pi^{2}}{9}

a practice on surd manipulation :biglaugh: \int_{\frac{1}{2}}^{1}\frac{\sin^{-1}x}{2+\sqrt{1+x}+\sqrt{1-x}}dx=\sqrt{2}-\sqrt{6}+\frac{\left(-2+11\sqrt{2}+2\sqrt{3}-3\sqrt{6}\right)\pi}{12}+\ln\left(20-12\sqrt{2}+10\sqrt{3}-8\sqrt{6}\right)-\frac{\pi^{2}}{9}

Looking forward to your approach.:)