Integral of sin^n(x)/(sin(x)+cos(x)+1)

Integration, One-Variable Calculus

Problem. Given integer n \ge 1, show that

\displaystyle \int_0^{\pi/2} \frac{\sin^nx}{\sin x + \cos x + 1} \ dx=\begin{cases}\frac{1}{2}\left(\frac{1}{(n-1)a_{\frac{n-2}{2}}}- \sum_{k=1}^{n-1}\frac{(-1)^{k+1}}{k}+\ln 2\right) \ \ \ \ \ \ \  \text{if} \ n \ \text{is even} \\ \\ \frac{1}{2}\left(\frac{\pi}{2}a_{\frac{n-1}{2}}+\sum_{k=1}^{n-1}\frac{(-1)^{k+1}}{k}-\ln 2\right) \ \ \ \ \ \ \ \ \text{if} \ n \ \text{is odd}, \end{cases}

where \displaystyle a_m:=\frac{\binom{2m}{m}}{4^m} for integers m \ge 0.

Solution. Let

\displaystyle I_n:=\int_0^{\pi/2} \frac{\sin^nx}{\sin x + \cos x + 1} \ dx

and see that

\displaystyle \begin{aligned}2I_n=\int_0^{\pi/2}\frac{(\sin x + \cos x - 1)\sin^nx}{\sin x \cos x} \ dx=\int_0^{\pi/2}\sin^{n-1}x \ dx - \int_0^{\pi/2} \frac{\sin^{n-1}x - \sin^nx}{\cos x} \ dx.\end{aligned}

So if we let

\displaystyle J_n:=\int_0^{\pi/2}\sin^{n-1}x \ dx, \ \ \ \ K_n:=\int_0^{\pi/2} \frac{\sin^{n-1}x - \sin^nx}{\cos x} \ dx,

then

\displaystyle 2I_n=J_n - K_n. \ \ \ \ \ \ \ (1)

We showed here that \displaystyle \int_0^{\pi/2} \sin^{2m}x \ dx = \frac{\pi}{2}a_m and \displaystyle \int_0^{\pi/2} \sin^{2m+1}x \ dx=\frac{1}{(2m+1)a_m}. So

\displaystyle J_n=\begin{cases} \frac{1}{(n-1)a_{\frac{n-2}{2}}} \ \ \ \ \ \ \text{if} \ n \ \text{is even} \\ \frac{\pi}{2}a_{\frac{n-1}{2}} \ \ \ \ \ \ \ \text{if} \ n \ \text{is odd}. \end{cases} \ \ \ \ \ \ \ (2)

We also have

\displaystyle \begin{aligned}K_n=\int_0^{\pi/2} \frac{\sin^{n-1}x(1-\sin x)}{\cos x} \ dx=\int_0^{\pi/2} \frac{\sin^{n-1}x \cos x}{1+\sin x} \ dx = \int_0^1 \frac{y^{n-1}}{1+y} \ dy, \ \ \ \ \ y=\sin x,\end{aligned}

\displaystyle =\int_0^1 \left(y^{n-2}-y^{n-3} + \cdots + (-1)^n + \frac{(-1)^{n+1}}{y+1}\right)dy

\displaystyle \begin{aligned}=\frac{1}{n-1}-\frac{1}{n-2}+\cdots + (-1)^n + (-1)^{n+1}\ln 2=(-1)^n\sum_{k=1}^{n-1}\frac{(-1)^{k+1}}{k}+(-1)^{n+1}\ln 2.\end{aligned} \ \ \ \ \ \ \ \ \ (3)

The result now follows from (1), (2), (3). \ \Box

Example.

\displaystyle \begin{aligned}\int_0^{\pi/2} \frac{\sin^9x+\cos^9x}{\sin x + \cos x + 1} \ dx=2\int_0^{\pi/2}\frac{\sin^9x}{\sin x+\cos x+1} \ dx=\frac{\pi}{2}a_4 + \sum_{k=1}^8 \frac{(-1)^{k+1}}{k}-\ln 2\end{aligned}

\displaystyle =\frac{35 \pi}{256}+\frac{533}{840}-\ln 2.

Published by Y. Sharifi

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