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Evaluation of $\displaystyle \int^{\frac{\pi}{2}}_{0}\frac{\cos^6 x}{\sin x+\cos x}dx$

$\bf{My\; Try::}$ Let $\displaystyle I = \int^{\frac{\pi}{2}}_{0}\frac{\cos^6 x}{\sin x+\cos x}dx = \int^{\frac{\pi}{2}}_{0}\frac{\sin^6 x}{\sin x+\cos x}dx$

Above we have used $\displaystyle \int^{a}_{0}f(x)dx = \int^{a}_{0}f(a-x)dx$

So $$2I = \int^{\frac{\pi}{2}}_{0}\frac{\sin^6 x+\cos^6 x}{\sin x+\cos x}dx = \int^{\frac{\pi}{2}}_{0}\frac{(\sin^2 x+\cos^2 x)(\sin^4 x+\cos^4 x-\sin^2 x\cos^2 x)}{\sin x+\cos x}dx$$

So $$2I = \int^{\frac{\pi}{2}}_{0}\frac{\sin^4 x+\cos^4 x-\sin^2 x\cos^2 x}{\sin x+\cos x}dx = \int^{\frac{\pi}{2}}_{0}\frac{1-3\sin^2 x\cos^2 x}{\sin x+\cos x}dx$$

Now How can i solve it after that, Help Required, Thanks

juantheron
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2 Answers2

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A little bit expansion of my above comment. Use $\cos x+\sin x=\sqrt{2}\cos(x-\pi/4) $ and changing variable, we get $$I=\int_{-\pi/4}^{\pi/4}\frac{\cos^6(x+\pi/4)}{\sqrt{2}\cos (x)}dx.$$ Since $\cos(x+\pi/4)=\frac{1}{\sqrt{2}}(\cos x-\sin x)$, one has $$I=\int_{-\pi/4}^{\pi/4}\frac{(\cos x-\sin x)^6}{(\sqrt 2)^7 \cos x}dx.$$

The following is easy.

Q-Zhang
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  • Please help with the final steps. Many thanks :-) – Abhimanyu Arora Sep 16 '16 at 16:05
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    @AbhimanyuArora Expand the numerator, remove the terms in odd sine power, for the other terms, linearize the trigs and integrate, except the last in $\sin^6x/\cos x$. For this one, write $\sin^2=1-\cos^2x$ and expand. You are left with a $1/\cos x$ to integrate, for which you get $\log(\tan(x/2+\pi/4))$. Straightforward, but a bit long (but not significantly longer than the other answer I think). – Jean-Claude Arbaut Sep 16 '16 at 18:14
  • @Jean-Claude: Merci, I still need to get the final integration of 1/cosx. What is the substitution you make there, t=tan(x/2)? – Abhimanyu Arora Sep 16 '16 at 23:53
  • Integral of sec(x) is $\ln|\sec x+ \tan x| $. – Q-Zhang Sep 17 '16 at 03:34
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    @AbhimanyuArora It's also $\ln|\tan(x/2+\pi/4)|$ (I forgot the absolute value in my preceding comment). See this for Qing Zhang's formula, and to see it's the same, write $\tan(x/2+\pi/4)=\dfrac{\cos(x/2)+\sin(x/2)}{\cos(x/2)-\sin(x/2)}$, then multiply the numerator and denominator by $\cos(x/2)+\sin(x/2)$, and you get $\dfrac{1+\sin(x)}{\cos(x)}=\tan(x)+\sec(x)$. – Jean-Claude Arbaut Sep 17 '16 at 06:03
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    @AbhimanyuArora Another way: to integrate $\dfrac{1}{\sin(x)}=\dfrac{1}{2\sin(x/2)\cos(x/2)}$, divide numerator and denominator by $\cos^2(x/2)$ to see it's equal to $\dfrac{1/(2\cos^2(x/2))}{\tan(x/2)}$, which you recognize as $\dfrac{f'(x)}{f(x)}$ with $f(x)=\tan(x/2)$, thus a primitive of $\dfrac{1}{\sin(x)}$ is $\log|\tan(x/2)|$. To integrate $\dfrac{1}{\cos(x)}$, notice $\dfrac{1}{\sin(x+\pi/2)}=\dfrac{1}{\cos(x)}$, thus a primitive of $\dfrac{1}{\cos(x)}$ is $\log|\tan((x+\pi/2)/2)|=\log|\tan(x/2+\pi/4)|$. – Jean-Claude Arbaut Sep 17 '16 at 06:39
2

Let $$　A=\int^{\frac{\pi}{2}}_{0}\frac{\cos^6 x}{\sin x+\cos x}dx, B=\int^{\frac{\pi}{2}}_{0}\frac{\sin^6 x}{\sin x+\cos x}dx. $$ Clearly $A=B$ by changing variable from $x$ to $\frac{\pi}{2}-x$. Now \begin{eqnarray} A+B&=&\int^{\frac{\pi}{2}}_{0}\frac{\cos^6 x+\sin^6x}{\sin x+\cos x}dx\\ &=&\int^{\frac{\pi}{2}}_{0}\frac{(\cos^2 x+\sin^2x)(\cos^4-\cos^2x\sin ^2x+\sin^4x)}{\sin x+\cos x}dx\\ &=&\int^{\frac{\pi}{2}}_{0}\frac{1-3\cos^2x\sin ^2x}{\sin x+\cos x}dx\\ &=&\int^{\frac{\pi}{2}}_{0}\frac{1-\frac34\sin ^2(2x)}{\sqrt{2}\sin(x+\frac{\pi}{4})}dx\\ &=&\int^{\frac{3\pi}{24}}_{\frac{\pi}{4}}\frac{1-\frac34\sin ^2[2(u-\frac{\pi}{4})]}{\sqrt{2}\sin u}du\\ &=&\int^{\frac{3\pi}{24}}_{\frac{\pi}{4}}\frac{1-\frac34\cos ^2(2u)}{\sqrt{2}\sin u}du\\ &=&\frac1{4\sqrt2}\int^{\frac{3\pi}{4}}_{\frac{\pi}{4}}\frac{4-3\cos ^2(2u)}{\sin u}du\\ &=&\frac1{4\sqrt2}\int^{\frac{3\pi}{4}}_{\frac{\pi}{4}}\frac{4-3(1-2\sin ^2u)^2}{\sin u}du\\ &=&\frac1{4\sqrt2}\int^{\frac{3\pi}{4}}_{\frac{\pi}{4}}(\frac{1}{\sin u}+12\sin ^2u-12\sin^4 u)du\\ &=&\frac1{4\sqrt2}\left[\ln\tan(\frac{u}{2})-3\cos u-3\cos (3u)\right]\bigg|^{\frac{3\pi}{4}}_{\frac{\pi}{4}}\\ &=&\frac12+\frac1{2\sqrt{2}}\ln(\cot(\frac{\pi}{8}))\\ &=&\frac12+\frac1{2\sqrt{2}}\coth^{-1}(\frac{1}{\sqrt2}) \end{eqnarray} and hence $$ A=B=\frac14+\frac1{4\sqrt{2}}\coth^{-1}(\frac{1}{\sqrt2}). $$

xpaul
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