【積分】King Property【置換積分】

$$P=\int_{0}^{\frac{\pi}{2}}\frac{sinx}{sinx+cosx}dx$$

を求めよ

$$P=\int_{0}^{\frac{\pi}{2}}\frac{sinx}{sinx+cosx}dx\cdots①$$

$x=\frac{\pi}{2}-\theta$とおくと

$$P=\int_{\frac{\pi}{2}}^{0}\frac{sin(\frac{\pi}{2}-\theta)}{sin(\frac{\pi}{2}-\theta)+cos(\frac{\pi}{2}-\theta)}(-d\theta)$$

$$P=\int_{\frac{\pi}{2}}^{0}\frac{cos\theta}{sin\theta+cos\theta}(-d\theta)$$

$$P=\int_{0}^{\frac{\pi}{2}}\frac{cos\theta}{sin\theta+cos\theta}d\theta\cdots②$$

①＋②をすると

$$2P=\int_{0}^{\frac{\pi}{2}}\frac{\sin\theta+cos\theta}{sin\theta+cos\theta}d\theta$$

約分して

$$2P=\int_{0}^{\frac{\pi}{2}}1\quad d\theta$$

$$2P=\left[\theta\right]_{0}^{\frac{\pi}{2}}$$

$$2P=\frac{\pi}{2}$$

$$P=\frac{\pi}{4}$$

$$P=\int_{0}^{\frac{\pi}{2}}\frac{sin^2x}{sinx+cosx}dx$$

を求めよ

$$P=\int_{0}^{\frac{\pi}{2}}\frac{sin^2x}{sinx+cosx}dx\cdots①$$

$x=\frac{\pi}{2}-\theta$とおくと

$$P=\int_{\frac{\pi}{2}}^{0}\frac{sin^2(\frac{\pi}{2}-\theta)}{sin(\frac{\pi}{2}-\theta)+cos(\frac{\pi}{2}-\theta)}(-d\theta)$$

$$P=\int_{\frac{\pi}{2}}^{0}\frac{cos^2\theta}{sin\theta+cos\theta}(-d\theta)$$

$$P=\int_{0}^{\frac{\pi}{2}}\frac{cos^2\theta}{sin\theta+cos\theta}d\theta\cdots②$$

①＋②をすると

$$2P=\int_{0}^{\frac{\pi}{2}}\frac{\sin^2\theta+cos^2\theta}{sin\theta+cos\theta}d\theta$$

$$2P=\int_{0}^{\frac{\pi}{2}}\frac{1}{sin\theta+cos\theta}d\theta$$

$$2P=\int_{0}^{\frac{\pi}{2}}\frac{1}{\sqrt{2}{sin(\theta+\frac{\pi}{4})}}d\theta$$

$$2P=\frac{1}{\sqrt{2}}\int_{0}^{\frac{\pi}{2}}{\frac{1}{sin(\theta+\frac{\pi}{4})}}d\theta$$

$t=\theta+\frac{\pi}{4}$とおくと

$$2P=\frac{1}{\sqrt{2}}\int_{\frac{\pi}{4}}^{\frac{3}{4}\pi}{\frac{1}{sint}}d\theta$$

$$2P=\frac{1}{\sqrt{2}}\int_{\frac{\pi}{4}}^{\frac{3}{4}\pi}{\frac{sinx}{sin^2t}}d\theta$$

$$2P=\frac{1}{\sqrt{2}}\int_{\frac{\pi}{4}}^{\frac{3}{4}\pi}{\frac{sinx}{1-cos^2t}}d\theta$$

$$2P=\frac{1}{\sqrt{2}}\int_{\frac{\pi}{4}}^{\frac{3}{4}\pi}{\frac{sinx}{(1-cost)(1+cost)}}d\theta$$

$$2P=\frac{1}{\sqrt{2}}\frac{1}{2}\int_{\frac{\pi}{4}}^{\frac{3}{4}\pi}\frac{sinx}{1-cost}+\frac{sinx}{1+cost}d\theta$$

$$2P=\frac{1}{2\sqrt{2}}\left[\log{(1-cost)}-\log{(1+cost)}\right]_{\frac{\pi}{4}}^{\frac{3}{4}\pi}$$

$$2P=\frac{1}{2\sqrt{2}}\left[\log{\frac{(1-cost)}{(1+cost)}}\right]_{\frac{\pi}{4}}^{\frac{3}{4}\pi}$$

$$2P=\frac{1}{2\sqrt{2}}\left[ \log{\frac{(1+\frac{1}{\sqrt2})}{(1-\frac{1}{\sqrt2})}}-\log{\frac{(1-\frac{1}{\sqrt2})}{(1+\frac{1}{\sqrt2})}}\right]$$

$$2P=\frac{1}{2\sqrt{2}} \log{\frac{(\sqrt2+1)^2}{(\sqrt2-1)^2}}$$

$$2P=\frac{1}{2\sqrt{2}} \log({\frac{\sqrt2+1}{\sqrt2-1}})^2$$

$$2P=\frac{1}{\sqrt{2}} \log({\frac{\sqrt2+1}{\sqrt2-1}})$$

$$2P=\frac{1}{\sqrt{2}} \log(\sqrt2+1)^2$$

$$2P=\frac{2}{\sqrt{2}}\log(\sqrt2+1)$$

よって

$$P=\frac{1}{\sqrt{2}} \log(\sqrt2+1)$$