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Find $$\int\frac{\sin^4 x+\cos^4 x}{\sin^3 x+\cos^3 x}dx$$

What I tried:

$$\sin^4(x)+\cos^4(x)=(\sin^2 x+\cos^2 x)^2-2\sin^2 x\cos^2 x=1-2\sin^2 x\cos^2 x$$

and $$\sin^3 x+\cos^3 x=(\sin x+\cos x)(1-\sin x\cos x)$$

so $$\int\frac{1-\sin^2 x\cos^2 x}{(\sin x+\cos x)(1-\sin x\cos x)}dx-\int\frac{\sin^2 x\cos^2 x}{(\sin x+\cos x)(1-\sin x\cos x)}dx$$

How do I solve it? Help me, please.

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    $\begingroup$ notice that $-2\sin^{2} x \cos^{2} x = \frac{-\sin ^{2} {2x}}{2}$ $\endgroup$ – Jneven Feb 21 at 14:25
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    $\begingroup$ Try the tan-half angle substitution $\endgroup$ – Dr. Sonnhard Graubner Feb 21 at 14:42
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    $\begingroup$ $$(a^3+b^3)(a+b)=a^4+b^4+ab(a^2+b^2)$$ We have $\dfrac{\sin^4x+\cos^4x}{\sin^3x+\cos^3x}=\sin x+\cos x-\dfrac{\sin x\cos x}{\sin^3x+\cos^3x}$ $\endgroup$ – lab bhattacharjee Feb 21 at 15:06
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A brutal way is just to enforce the substitution $x=2\arctan t$, leading to $$ \int\frac{\sin x\cos x}{\sin^3 x+\cos^3 x}\,dx=2\int\frac{\frac{2t(1-t^2)}{(1+t^2)^3}}{\frac{(2t)^3}{(1+t^2)^3}+\frac{(1-t^2)^3}{(1+t^2)^3}}\,dt=\int\frac{4t(1-t^2)}{(2t)^3+(1-t^2)^3}\,dt $$ an ugly integral, but perfectly solvable by partial fraction decomposition. The roots of $(2t)^3+(1-t^2)^3$ can be found by solving the quadratic equations given by $$\frac{1-t^2}{2t}\in\left\{-1,\frac{1+i\sqrt{3}}{2},\frac{1-i\sqrt{3}}{2}\right\}.$$

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