Trick, or Indefinite Integration?

Question

Evaluate $$\int_{0}^{\pi/2}\dfrac{\sin^6 x}{\sin x + \cos x}\text{ d}x$$

Is there a nice, elegant way of solving the above integral?

Here's what I did - Replaced $f(x)$ by $f(\pi/2-x)$, added and got rid of some common factors in numerator and denominator. This was followed by indefinite integration, and ultimately substituting the limits in the final expression.

The above method is kind of long, though - and I'm wondering if there's a quicker, shorter and rather elegant way of getting around this integral. Could someone post a solution, and share ideas?

Thanks a lot.

Answer 1

A standard way for dealing with trigonometric integrals is to exploit the substitution $x=2\arctan\frac{t}{2}$.
Here it leads to $$ \int_{0}^{\pi/2}\frac{\sin^6 x}{\sin x+\cos x}\,dx = 128\int_{0}^{1}\frac{t^6}{(2t+1-t^2)(t^2+1)^6}\,dt$$ and the last integral can be computed by partial fraction decomposition. It equals $$ \frac{1}{8}\left(2+\sqrt{2}\,\text{arctanh}\frac{1}{\sqrt{2}}\right)\approx\frac{28}{69}. $$ This also follows from $\sin x+\cos x=\sqrt{2}\cos\left(x-\frac{\pi}{4}\right)$ and $$ \int_{-\pi/4}^{\pi/4}\frac{\sin^6\left(x+\frac{\pi}{4}\right)}{\sqrt{2}\cos x}\,dx=\frac{1}{8\sqrt{2}}\sum_{k=0}^{6}\binom{6}{k}\int_{-\pi/4}^{\pi/4}\sin^k(x)\cos^{5-k}(x)\,dx $$ where the contribution provided by odd $k$s is zero and $$ 2\int_{0}^{\pi/4}\cos(x)^5\,dx =\frac{43}{30\sqrt{2}},\qquad 2\int_{0}^{\pi/4}\sin(x)^2\cos(x)^3\,dx =\frac{7}{30\sqrt{2}}$$ $$ 2\int_{0}^{\pi/4}\sin(x)^4\cos(x)^1\,dx =\frac{3}{30\sqrt{2}}$$ $$2\int_{0}^{\pi/4}\sin(x)^6\cos(x)^{-1}\,dx =-\frac{73}{30\sqrt{2}}+4\,\text{arctanh}\tan\frac{\pi}{8}.$$

Answer 2

I have solved it first as a indefinite integral and then as a definite integral. I am not sure how different it is with yours. Good luck. Let me know if you have any questions.

Answer 3

Let $J$ be the integral to be calculated, we want to get the computer result, here using sage:

sage: var('x');
sage: J = integral( sin(x)^6/(sin(x)+cos(x)), x, 0, pi/2 )
sage: J
1/16*sqrt(2)*log((2*sqrt(2)) + 3) + 1/4

For this, we use the simple idea of getting a $\sin(x+\pi/4)$ (up to a multiplicative constant) in the denominator, then do trivial computations...: $$ \begin{aligned} J &=\int _0^{\pi/2} \frac{\sin^6 x}{\sin x + \cos x}\;dx \\ &=\int _0^{\pi/2}\frac 1{\sqrt 2} \frac{\sin^6 x}{\frac 1{\sqrt 2}\left(\sin x + \cos x\right)}\;dx \\ &=\frac 1{\sqrt 2} \int _0^{\pi/2} \frac {\sin^6 x} {\sin \left(x + \frac\pi4\right)}\;dx \\ &\qquad\text{Substitution: }y = x+\frac \pi 4\ , \\ &=\frac 1{\sqrt 2} \int_{\pi/4}^{3\pi/4} \frac {\sin^6 \left(y- \frac\pi4\right)} {\sin y}\;dy \\ &=\frac 1{\sqrt 2} \int_{\pi/4}^{3\pi/4} \frac 18\cdot\frac 1{\sin y} \Big(\ \sin y-\cos y\ \Big)^6\;dy \\ &=\frac 1{\sqrt 2} \int_{\pi/4}^{3\pi/4} \frac 18\cdot\frac 1{\sin y} \Big(\ \sin^6 y -6\sin^5y\cos y +15\sin^4y\cos^2 y \\&\qquad\qquad -20\sin^3y\cos^3 y +15\sin^2y\cos^4 y -6\sin y\cos^5 y +\cos^6 y \ \Big)\;dy \\ &=\frac 1{8\sqrt 2} \int_{\pi/4}^{3\pi/4} \Big(\ \sin^5 y -6\sin^4y\cos y +15\sin^3y\cos^2 y \\&\qquad\qquad -20\sin^2y\cos^3 y +15\sin y\cos^4 y -6\cos^5 y +\frac{\cos^6 y}{\sin y} \ \Big)\;dy \\ &= \frac 1{8\sqrt 2} \int_{\pi/4}^{3\pi/4}\left[\ \begin{aligned} -(1-\cos^2y)^2 &\,\cdot\,\cos 'y \\ - 6\sin^4y &\,\cdot\,\sin'y \\ - 15(1-\cos^2y)\cos^2y &\,\cdot\,\cos'y \\ - 20\sin^2y(1-\sin^2 y) &\,\cdot\,\sin'y \\ - 15\cos^4y &\,\cdot\,\cos'y \\ - 6(1-\sin^2 y)^2 &\,\cdot\,\sin'y \\ -\frac{\cos ^6y}{1-\cos^2 y}&\,\cdot\,\cos'y \end{aligned} \ \right]\; dy \\ &= \frac 1{8\sqrt 2} \int_{\pi/4}^{3\pi/4}\left[\ \begin{aligned} -(1-\cos^2y)^2 & \\ - 15(1-\cos^2y)\cos^2y & \\ - 15\cos^4y & \\ -\frac{\cos ^6y}{1-\cos^2 y}& \end{aligned} \ \right]\,\cdot\,\cos 'y\; dy \\ &\qquad\text{because for the $\sin'y$-terms we have $\sin(\pi/4)=\sin(3\pi/4)$...} \\ &\qquad\text{Substitution: }u=\cos t\ ,\ du=\cos't\, dt\ , \\ &= \frac 1{8\sqrt 2} \int_{-1/\sqrt2}^{1/\sqrt2}\left[\ (1-u^2)^2 +15(1-u^2)u^2 +15u^4 +\frac {u^6-1+1}{1-u^2} \right] \; du \\ &= \frac 1{4\sqrt 2} \int_0^{1/\sqrt2}\left[\ (1-u^2)^2 +15(1-u^2)u^2 +15u^4 -(1+u^2+u^4) +\frac {1}{1-u^2} \right] \; du \\ &= \frac 1{4\sqrt 2} \int_0^{1/\sqrt2}\left[\ 12u^2 +\frac {1}{1-u^2} \right] \; du \\ &= \frac 1{4\sqrt 2} \left[\ 4\left(\frac 1{\sqrt2}\right)^3 +\frac 12\log \frac{1+\frac 1{\sqrt 2}}{1-\frac 1{\sqrt 2}} \ \right] \\ &= \frac 14 + \frac 1{4\sqrt 2}\log(\sqrt2+1)) \ . \end{aligned} $$

Answer 4

Hint:

$$2I=\int_0^{\pi/2}\dfrac{\sin^6x+\cos^6x}{\sin x+\cos x}dx$$

Now $\sin^6x+\cos^6x=(\sin^2x+\cos^2x)^3-3\sin^2x\cos^2x(\sin^2x+\cos^2x)=1-3\sin^2x\cos^2x$

and $\sin^2x\cos^2x=\dfrac{\sin^22x}4=\dfrac{1-\cos4x}8$

For $J=\displaystyle\int_0^{\pi/2}\dfrac{\cos4x}{\sin x+\cos x}dx=\int_0^{\pi/2}\dfrac{\cos4x}{\sqrt2\cos\left(x-\dfrac\pi4\right)}dx,$

set $x-\dfrac\pi4=y,\cos4x=\cos4\left(y+\dfrac\pi4\right)=\cos(4y+\pi)=-\cos4y$

$\implies-\sqrt2J=\displaystyle\int_{-\pi/4}^{\pi/4}\dfrac{\cos4ydy}{\cos y}=\int_{-\pi/4}^{\pi/4}\dfrac{1-8\cos^2y+8\cos^4y}{\cos y}dy$

Now use $\cos3y=4\cos^3y-3\cos y $

Answer 5

Substitute $t=\sin x-\cos x$ to have $dt=\sin x+\cos x$ $$(\sin x+\cos x)^2= 2-t^2,\>\>\>\>\>\sin^6x+\cos^6x=\frac14( 1+6t^2-3t^4)$$ Then

\begin{align}\int_{0}^{\pi/2}\dfrac{\sin^6 x}{\sin x + \cos x}{ d}x =& \ \frac12 \int_{0}^{\pi/2}\dfrac{\sin^6 x+\cos^6 x}{\sin x + \cos x}{ d}x\\ =&\ \frac18 \int_{-1}^1\frac{1+6t^2-3t^4}{2-t^2}dt = \frac14\left(1+\frac{\coth^{-1}\sqrt2}{\sqrt2}\right) \end{align}