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Integrate $$I=\int\frac{\cos^2x}{1+\tan x}dx$$


$$I=\int\frac{\cos^3xdx}{\cos x+\sin x}=\int\frac{\cos^3x(\cos x-\sin x)dx}{\cos^2x-\sin^2x}=\int\frac{\cos^4xdx}{1-2\sin^2x}-\int\frac{\cos^3x\sin xdx}{2\cos^2x-1}$$ Let $t=\sin x,u=\cos x,dt=\cos xdx,du=-\sin xdx$ $$I=\underbrace{\int\frac{-u^4du}{(2u^2-1)\sqrt{1-u^2}}}_{I_1}+\underbrace{\int\frac{u^3du}{2u^2-1}}_{I_2}$$ I have found(using long division): $$I_2=\frac{u^2}2+\frac18\ln|2u^2-1|+c=\frac12\cos^2x+\frac18\ln|\cos2x|$$ I have converted $I_1$ into this: $$I_1=\frac12\left(\int(-2)\sqrt{1-u^2}du+\int\frac{du}{\sqrt{1-u^2}}\right)+\frac14\underbrace{\int\frac{du}{(2u^2-1)\sqrt{1-u^2}}}_{I_3}$$ Now I have took $v=1/u$ in $I_3$ so that $du=-(1/v^2)dv$: $$I_3=\int\frac{vdv}{(v^2-2)\sqrt{v^2-1}}$$ Now I took $w^2=v^2-1$ or $wdw=vdv$ to get: $$I_3=\int\frac{dw}{w^2-1}=\frac12\ln\left|\frac{w-1}{w+1}\right|$$ I have not yet formulated the entire thing;


  • Is this correct?
  • This is very long, do you have any "shorter" method?
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We have $$I=\int\dfrac{\cos^3x}{\cos x+\sin x}dx=\dfrac1{\sqrt2}\int\dfrac{\cos^3x}{\cos\left(\dfrac\pi4-x\right)}dx$$

Setting $\dfrac\pi4-x=y\iff x=\dfrac\pi4-y,dx=-dy$

$$-\sqrt2I=\int\dfrac{\cos^3\left(\dfrac\pi4-y\right)}{\cos y}dy=\dfrac1{2\sqrt2}\int\dfrac{(\cos y+\sin y)^3}{\cos y}dy$$

The rest is pretty easy.

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  • $\begingroup$ is my procedure correct? $\endgroup$ – RE60K Sep 14 '14 at 9:00
  • $\begingroup$ the rest I can do from your answer, don't keep adding little $\endgroup$ – RE60K Sep 14 '14 at 9:01
  • $\begingroup$ how would you proceed from there (just a rough method)? Sorry I don't see it :(. expand? or is there some sneaky trick? $\endgroup$ – surelyourejoking Sep 14 '14 at 9:12
  • $\begingroup$ @surelyourejoking expand and then take various substitutions $\tan x$ is universal but several cases involve $\cos x$ and $\sin x$ depending upon the integral. $\endgroup$ – RE60K Sep 14 '14 at 9:14
  • $\begingroup$ @Aditya Ok sounds good, thanks! $\endgroup$ – surelyourejoking Sep 14 '14 at 9:19
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The integral is equivalent to $ \displaystyle{\int\frac{\cos^3 x}{\sin x + \cos x}\,\mathrm{d}x}. $ Now, consider $$ \mathcal{I}_{1} = \int\frac{\cos^3 x}{\sin x + \cos x}\,\mathrm{d}x \qquad\text{and}\qquad\mathcal{I}_{2}=\int\frac{\sin^3 x}{\sin x + \cos x}\,\mathrm{d}x. $$ Observe that $$ \begin{aligned} \mathcal{I}_{1}+\mathcal{I}_{2} &= \int\frac{\cos^3 x + \sin^3 x}{\sin x + \cos x}\,\mathrm{d}x\\ &=\int\frac{(\cos x + \sin x)(\cos^2 x - \sin x\cos x + \sin^2 x)}{\cos x + \sin x}\,\mathrm{d}x\\ &=x - \frac{\sin^2 x}{2}+C_{1}, \end{aligned} $$ and $$ \begin{aligned} \mathcal{I}_{1}-\mathcal{I}_{2}&=\int\frac{(\cos x - \sin x)(1 + \sin x \cos x)}{\cos x + \sin x}\,\mathrm{d}x\\ &=\int\frac{(\cos x - \sin x)(\cos x + \sin x)(1 + \sin x\cos x)}{(\cos x + \sin x)^2}\,\mathrm{d}x\\ &=\int\frac{1 +\frac{\sin 2x}{2}}{1 + \sin 2x}\cos 2x\,\mathrm{d}x\\ &=\underbrace{\frac{1}{4}\int\left(1 + \frac{1}{1+\sin 2x}\right)2\cos 2x\,\mathrm{d}x}_{t = \sin 2x\implies\mathrm{d}t=2\cos 2x\,\mathrm{d}x}\\ &=\frac{1}{4}\int1+\frac{1}{1+t}\,\mathrm{d}t\\ &=\frac{t}{4}+\frac{1}{4}\log|t+1|+C_{2}\\ &=\frac{\sin2x}{4}+\frac{1}{4}\log(1+\sin2x)+C_{2} \end{aligned} $$. Thus, $$ \begin{aligned} (\mathcal{I}_{1}+\mathcal{I}_{2})+(\mathcal{I}_{1}-\mathcal{I}_{2})&=2\mathcal{I}_{1}\\ &=x+\frac{\sin2x}{4}+\frac{1}{4}\log(1+\sin2x)-\frac{\sin^2 x}{2} + C_{3}, \end{aligned} $$ which yields $$ \begin{aligned} \mathcal{I}_{1}&=\int\frac{\cos^3x}{\sin x + \cos x}\,\mathrm{d}x\\ &=\int\frac{\cos^2 x}{1+\tan x}\,\mathrm{d}x\\ &=\frac{x}{2}+\frac{\sin 2x}{8}+\frac{1}{8}\log(1+\sin2x)-\frac{\sin^2 x}{4} + C, \end{aligned} $$ or, in an equivalent manner, $$ \int\frac{\cos^2 x}{1+\tan x}\,\mathrm{d}x=\frac{x}{2}+\frac{\sin x\cos x}{4}+\frac{1}{4}\log\left|\cos\!\left(x-\frac{\pi}{4}\right)\right|-\frac{\sin^2 x}{4}+\mathcal{C}. $$

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Setting $\tan x=u,I=\int\dfrac{\sec^2x}{\sec^4x(1+\tan x)}dx$

$I=\int\dfrac{du}{(1+u^2)^2(1+u)}=\int\dfrac1{u(1+u)}\cdot\dfrac u{(1+u^2)^2}du$

Integrating by parts, $I=\dfrac1{u(1+u)}\int\dfrac u{(1+u^2)^2}\ du-\int\left(\dfrac{d\dfrac1{u(1+u)}}{du}\int\dfrac u{(1+u^2)^2}\ du\right)du$

$I=-\dfrac1{2u(1+u)(1+u^2)}-\int\dfrac{2u+1}{2u^2(1+u)^2(1+u^2)}du$

Now use Partial Fraction Decomposition,

$\dfrac{2u+1}{u^2(1+u)^2(1+u^2)}=\dfrac Au+\dfrac B{u^2}+\dfrac C{1+u}+\dfrac D{(1+u)^2}+\dfrac E{1+u^2}$

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  • $\begingroup$ surely I can't select two answers! $\endgroup$ – RE60K Sep 14 '14 at 9:22
  • $\begingroup$ does'n that should be $D\mapsto Du+D',E\mapsto Eu+E'$? $\endgroup$ – RE60K Sep 14 '14 at 9:25
  • $\begingroup$ @Aditya, I could not make out the last statement. I believe there exists even better solution! $\endgroup$ – lab bhattacharjee Sep 14 '14 at 9:27
  • $\begingroup$ 1)sorry that comment is wrong.2)yes there might one existing too> $\endgroup$ – RE60K Sep 14 '14 at 9:31

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