Let's do some calculus! (45)

Calculus Level 5

$\large \displaystyle \int \dfrac{\sin^4 x + \cos^4 x}{\sin^3 x + \cos^3 x} \, dx$

If the above integral can be represented as

$\sin x - \cos x + \frac{\sqrt{a}}{b} \cdot \operatorname{arctanh} \left( \frac{ \tan \left( \frac x2 \right) - 1 }{\sqrt{a}} \right) + \frac ab \cdot \arctan \left( \cos x - \sin x \right) + C,$

where $a$ and $b$ are coprime positive integers with $a$ square free, then evaluate $a+b$ .

Clarification: $C$ denotes the arbitrary constant of integration .

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The answer is 5.

1 solution

Mark Hennings
Apr 12, 2017

Note that $\begin{aligned} \frac{\sin^4x + \cos^4x}{\sin^3x + \cos^3x} & = \frac{(\sin x + \cos x)(\sin^3x + \cos^3x) - \sin x \cos x(\sin^2x + \cos^2x)}{\sin^3x + \cos^3x} \\ & = \sin x + \cos x - \frac{\sin x \cos x}{(\sin x + \cos x)(\sin^2x + \cos^2x) - \sin x \cos x(\sin x + \cos x)} \\ & = \sin x + \cos x - \frac{\sin x \cos x}{(\sin x + \cos x)(1 - \sin x \cos x)} \\ & = \sin x + \cos x + \tfrac13\left(\frac{1}{\sin x + \cos x} - \frac{\sin x + \cos x}{1 - \sin x \cos x}\right) \\ & = \sin x + \cos x - \tfrac23 \frac{\sin x + \cos x}{1 + (\cos x - \sin x)^2} + \frac{1}{3(\sin x + \cos x)} \end{aligned}$ and so $\int \frac{\sin^4x + \cos^4x}{\sin^3x + \cos^3x}\,dx \; = \; \sin x - \cos x +\tfrac23\tan^{-1}\big(\cos x - \sin x) + \int \frac{dx}{3(\sin x + \cos x)}$ Now the substitution $t = \tan\tfrac12x$ yields $\begin{aligned} \int \frac{dx}{\sin x + \cos x} & = \int \frac{1 + t^2}{2t + (1-t^2)}\, \frac{2\,dt}{1+t^2} \; = \; 2\int \frac{dt}{1 + 2t - t^2} \\ & = 2\int \frac{dt}{2 - (t-1)^2} \; = \; \sqrt{2}\tanh^{-1}\big(\tfrac{1}{\sqrt2}(t-1)\big) + c \end{aligned}$ and hence $\int \frac{\sin^4x + \cos^4x}{\sin^3x + \cos^3x}\,dx \; = \; \sin x - \cos x + \tfrac23\tan^{-1}\big(\cos x - \sin x) + \frac{\sqrt{2}}{3}\tanh^{-1}\Big(\frac{\tan(\tfrac12x)-1}{\sqrt{2}}\Big) + C$ so that $a=2,b=3$ , and the answer is $\boxed{5}$ .

Most of the terms in the expression to be derived already give one a hint about which functions to pick. Maybe I should have omitted some more key hints here. Don't you think?

P.S. Your solution is as great as always sir!

Tapas Mazumdar - 4 years, 1 month ago

Yes the term $sinx-cosx$ and $arctan(cosx-sinx)$ gives a hint of what is to be substituted.You should have tried to change to some other form just for confusion.

Spandan Senapati - 4 years, 1 month ago

Yes exactly! Sad I noticed it later after I had already uploaded the problem. Still it's up level 5 so I guess it's good enough.