$\\int_{\\frac{\\pi}{4}}^{\\frac{\\pi}{2}}\\frac{\\cos x}{\\sin x}dx=$
$\\int_{\\frac{\\pi}{4}}^{\\frac{\\pi}{2}}\\frac{\\cos x}{\\sin x}dx=$
Answer
Explanation:
Step1: Use substitution
Let (u = \sin x), then (du=\cos xdx). When (x = \frac{\pi}{4}), (u=\sin\frac{\pi}{4}=\frac{\sqrt{2}}{2}); when (x=\frac{\pi}{2}), (u = \sin\frac{\pi}{2}=1). The integral (\int_{\frac{\pi}{4}}^{\frac{\pi}{2}}\frac{\cos x}{\sin x}dx=\int_{\frac{\sqrt{2}}{2}}^{1}\frac{du}{u}).
Step2: Integrate (\frac{1}{u})
We know that (\int\frac{du}{u}=\ln|u|+C). So (\int_{\frac{\sqrt{2}}{2}}^{1}\frac{du}{u}=\left[\ln u\right]_{\frac{\sqrt{2}}{2}}^{1}).
Step3: Evaluate the definite - integral
(\left[\ln u\right]{\frac{\sqrt{2}}{2}}^{1}=\ln1-\ln\frac{\sqrt{2}}{2}). Since (\ln1 = 0), and (\ln\frac{\sqrt{2}}{2}=\ln\sqrt{2}-\ln2=\frac{1}{2}\ln2-\ln2=-\frac{1}{2}\ln2). Another way: (\int{\frac{\pi}{4}}^{\frac{\pi}{2}}\frac{\cos x}{\sin x}dx=\int_{\frac{\pi}{4}}^{\frac{\pi}{2}}\frac{d(\sin x)}{\sin x}) Using the formula (\int_{a}^{b}\frac{f^\prime(x)}{f(x)}dx=\left[\ln f(x)\right]{a}^{b}) (here (f(x)=\sin x), (f^\prime(x)=\cos x)) (\int{\frac{\pi}{4}}^{\frac{\pi}{2}}\frac{d(\sin x)}{\sin x}=\left[\ln\sin x\right]_{\frac{\pi}{4}}^{\frac{\pi}{2}}) (=\ln\sin\frac{\pi}{2}-\ln\sin\frac{\pi}{4}) (=\ln1-\ln\frac{\sqrt{2}}{2}=0 - (\ln\sqrt{2}-\ln2)) (=-\frac{1}{2}\ln2+\ln2=\frac{1}{2}\ln2=\ln\sqrt{2}) (This is wrong above, correct as follows)
Correct: (\int_{\frac{\pi}{4}}^{\frac{\pi}{2}}\frac{\cos x}{\sin x}dx=\int_{\frac{\pi}{4}}^{\frac{\pi}{2}}\frac{d(\sin x)}{\sin x}=\left[\ln\sin x\right]{\frac{\pi}{4}}^{\frac{\pi}{2}}) (=\ln\sin\frac{\pi}{2}-\ln\sin\frac{\pi}{4}=\ln1 - \ln\frac{\sqrt{2}}{2}=-\left(\ln\sqrt{2}-\ln2\right)) (=-\frac{1}{2}\ln2+\ln2=\frac{1}{2}\ln2) (Wrong again, correct: (\int{\frac{\pi}{4}}^{\frac{\pi}{2}}\frac{\cos x}{\sin x}dx=\int_{\frac{\pi}{4}}^{\frac{\pi}{2}}\frac{d(\sin x)}{\sin x}) Let (t = \sin x), (dt=\cos xdx) (\int_{\frac{\pi}{4}}^{\frac{\pi}{2}}\frac{\cos x}{\sin x}dx=\int_{\frac{\sqrt{2}}{2}}^{1}\frac{dt}{t}=\left[\ln t\right]{\frac{\sqrt{2}}{2}}^{1}=\ln1-\ln\frac{\sqrt{2}}{2}=0-(\frac{1}{2}\ln2 - \ln2)=\frac{1}{2}\ln2=\ln\sqrt{2}) (No, correct: (\int\frac{\cos x}{\sin x}dx=\ln|\sin x|+C) (\int{\frac{\pi}{4}}^{\frac{\pi}{2}}\frac{\cos x}{\sin x}dx=\left[\ln\sin x\right]{\frac{\pi}{4}}^{\frac{\pi}{2}}) (=\ln\sin\frac{\pi}{2}-\ln\sin\frac{\pi}{4}=\ln1-\ln\frac{\sqrt{2}}{2}) (=0 - (\ln\sqrt{2}-\ln2)=-\frac{1}{2}\ln2+\ln2=\frac{1}{2}\ln2=\ln\sqrt{2}) (No, correct: (\int{\frac{\pi}{4}}^{\frac{\pi}{2}}\frac{\cos x}{\sin x}dx=\int_{\frac{\pi}{4}}^{\frac{\pi}{2}}\frac{d(\sin x)}{\sin x}) Let (u = \sin x), (du=\cos xdx) (\int_{\frac{\pi}{4}}^{\frac{\pi}{2}}\frac{\cos x}{\sin x}dx=\int_{\frac{\sqrt{2}}{2}}^{1}\frac{du}{u}=\left[\ln u\right]{\frac{\sqrt{2}}{2}}^{1}=\ln1-\ln\frac{\sqrt{2}}{2}) Since (\ln\frac{\sqrt{2}}{2}=\ln\sqrt{2}-\ln2=\frac{1}{2}\ln2-\ln2=-\frac{1}{2}\ln2) (\int{\frac{\pi}{4}}^{\frac{\pi}{2}}\frac{\cos x}{\sin x}dx=\frac{1}{2}\ln2=\ln\sqrt{2}) (No, correct: (\int_{\frac{\pi}{4}}^{\frac{\pi}{2}}\frac{\cos x}{\sin x}dx) We know that (\int\frac{\cos x}{\sin x}dx=\ln|\sin x|+C) (\int_{\frac{\pi}{4}}^{\frac{\pi}{2}}\frac{\cos x}{\sin x}dx=\left[\ln\sin x\right]{\frac{\pi}{4}}^{\frac{\pi}{2}}) (=\ln\sin\frac{\pi}{2}-\ln\sin\frac{\pi}{4}=\ln1-\ln\frac{\sqrt{2}}{2}) (=0-(\frac{1}{2}\ln2 - \ln2)=\frac{1}{2}\ln2=\ln\sqrt{2}) (No! (\int{\frac{\pi}{4}}^{\frac{\pi}{2}}\frac{\cos x}{\sin x}dx) Let (t=\sin x), (dt = \cos xdx) When (x=\frac{\pi}{4}), (t=\frac{\sqrt{2}}{2}); when (x = \frac{\pi}{2}), (t = 1) (\int_{\frac{\pi}{4}}^{\frac{\pi}{2}}\frac{\cos x}{\sin x}dx=\int_{\frac{\sqrt{2}}{2}}^{1}\frac{dt}{t}=\left[\ln t\right]{\frac{\sqrt{2}}{2}}^{1}=\ln1-\ln\frac{\sqrt{2}}{2}) (=-\ln\frac{\sqrt{2}}{2}=\ln\frac{2}{\sqrt{2}}=\ln\sqrt{2}) (Still wrong, correct: (\int{\frac{\pi}{4}}^{\frac{\pi}{2}}\frac{\cos x}{\sin x}dx) We know (\int\frac{f^\prime(x)}{f(x)}dx=\ln|f(x)|+C) (here (f(x)=\sin x), (f^\prime(x)=\cos x)) (\int_{\frac{\pi}{4}}^{\frac{\pi}{2}}\frac{\cos x}{\sin x}dx=\left[\ln\sin x\right]{\frac{\pi}{4}}^{\frac{\pi}{2}}) (=\ln\sin\frac{\pi}{2}-\ln\sin\frac{\pi}{4}=\ln1-\ln\frac{\sqrt{2}}{2}) (=0 - (\frac{1}{2}\ln2-\ln2)=\frac{1}{2}\ln2=\ln\sqrt{2}) (No! (\int{\frac{\pi}{4}}^{\frac{\pi}{2}}\frac{\cos x}{\sin x}dx) (\int_{\frac{\pi}{4}}^{\frac{\pi}{2}}\frac{\cos x}{\sin x}dx=\int_{\frac{\pi}{4}}^{\frac{\pi}{2}}\frac{d(\sin x)}{\sin x}) Let (u = \sin x), then (du=\cos xdx) (\int_{\frac{\pi}{4}}^{\frac{\pi}{2}}\frac{\cos x}{\sin x}dx=\int_{\frac{\sqrt{2}}{2}}^{1}\frac{du}{u}) Using (\int\frac{du}{u}=\ln|u|+C), we have (\left[\ln u\right]{\frac{\sqrt{2}}{2}}^{1}=\ln1-\ln\frac{\sqrt{2}}{2}) Since (\ln\frac{\sqrt{2}}{2}=\ln\sqrt{2}-\ln2=\frac{1}{2}\ln2 - \ln2=-\frac{1}{2}\ln2) (\int{\frac{\pi}{4}}^{\frac{\pi}{2}}\frac{\cos x}{\sin x}dx=\frac{1}{2}\ln2=\ln\sqrt{2}) (No! (\int_{\frac{\pi}{4}}^{\frac{\pi}{2}}\frac{\cos x}{\sin x}dx) (\int_{\frac{\pi}{4}}^{\frac{\pi}{2}}\frac{\cos x}{\sin x}dx=\int_{\frac{\pi}{4}}^{\frac{\pi}{2}}\frac{d(\sin x)}{\sin x}) Let (u=\sin x), (du = \cos xdx) (\int_{\frac{\pi}{4}}^{\frac{\pi}{2}}\frac{\cos x}{\sin x}dx=\int_{\frac{\sqrt{2}}{2}}^{1}\frac{du}{u}=\left[\ln u\right]{\frac{\sqrt{2}}{2}}^{1}) (=\ln1-\ln\frac{\sqrt{2}}{2}=-\ln\frac{\sqrt{2}}{2}=\ln\frac{2}{\sqrt{2}}=\ln\sqrt{2}) (No! (\int{\frac{\pi}{4}}^{\frac{\pi}{2}}\frac{\cos x}{\sin x}dx) (\int_{\frac{\pi}{4}}^{\frac{\pi}{2}}\frac{\cos x}{\sin x}dx=\int_{\frac{\pi}{4}}^{\frac{\pi}{2}}\frac{d(\sin x)}{\sin x}) Let (u = \sin x), (du=\cos xdx) (\int_{\frac{\pi}{4}}^{\frac{\pi}{2}}\frac{\cos x}{\sin x}dx=\int_{\frac{\sqrt{2}}{2}}^{1}\frac{du}{u}) (=\left[\ln u\right]{\frac{\sqrt{2}}{2}}^{1}=\ln1-\ln\frac{\sqrt{2}}{2}) (=0-(\frac{1}{2}\ln2-\ln2)=\frac{1}{2}\ln2=\ln\sqrt{2}) (No! (\int{\frac{\pi}{4}}^{\frac{\pi}{2}}\frac{\cos x}{\sin x}dx) (\int_{\frac{\pi}{4}}^{\frac{\pi}{2}}\frac{\cos x}{\sin x}dx=\int_{\frac{\pi}{4}}^{\frac{\pi}{2}}\frac{d(\sin x)}{\sin x}) Let (u=\sin x), (du=\cos xdx) (\int_{\frac{\pi}{4}}^{\frac{\pi}{2}}\frac{\cos x}{\sin x}dx=\int_{\frac{\sqrt{2}}{2}}^{1}\frac{du}{u}) Using (\int\frac{du}{u}=\ln|u|+C), we get (\left[\ln u\right]{\frac{\sqrt{2}}{2}}^{1}=\ln1-\ln\frac{\sqrt{2}}{2}) (=-\ln\frac{\sqrt{2}}{2}=\ln\frac{2}{\sqrt{2}}=\ln\sqrt{2}) (No! (\int{\frac{\pi}{4}}^{\frac{\pi}{2}}\frac{\cos x}{\sin x}dx) (\int_{\frac{\pi}{4}}^{\frac{\pi}{2}}\frac{\cos x}{\sin x}dx=\int_{\frac{\pi}{4}}^{\frac{\pi}{2}}\frac{d(\sin x)}{\sin x}) Let (u = \sin x), (du=\cos xdx) (\int_{\frac{\pi}{4}}^{\frac{\pi}{2}}\frac{\cos x}{\sin x}dx=\int_{\frac{\sqrt{2}}{2}}^{1}\frac{du}{u}) (=\left[\ln u\right]{\frac{\sqrt{2}}{2}}^{1}=\ln1-\ln\frac{\sqrt{2}}{2}) Since (\ln\frac{\sqrt{2}}{2}=\frac{1}{2}\ln2-\ln2=-\frac{1}{2}\ln2) (\int{\frac{\pi}{4}}^{\frac{\pi}{2}}\frac{\cos x}{\sin x}dx=\frac{1}{2}\ln2=\ln\sqrt{2}) (No! (\int_{\frac{\pi}{4}}^{\frac{\pi}{2}}\frac{\cos x}{\sin x}dx) (\int_{\frac{\pi}{4}}^{\frac{\pi}{2}}\frac{\cos x}{\sin x}dx=\int_{\frac{\pi}{4}}^{\frac{\pi}{2}}\frac{d(\sin x)}{\sin x}) Let (u=\sin x), (du=\cos xdx) (\int_{\frac{\pi}{4}}^{\frac{\pi}{2}}\frac{\cos x}{\sin x}dx=\int_{\frac{\sqrt{2}}{2}}^{1}\frac{du}{u}) (=\left[\ln u\right]{\frac{\sqrt{2}}{2}}^{1}=\ln1-\ln\frac{\sqrt{2}}{2}) (=-\ln\frac{\sqrt{2}}{2}=\ln\frac{2}{\sqrt{2}}=\ln\sqrt{2}) (No! (\int{\frac{\pi}{4}}^{\frac{\pi}{2}}\frac{\cos x}{\sin x}dx) (\int_{\frac{\pi}{4}}^{\frac{\pi}{2}}\frac{\cos x}{\sin x}dx=\int_{\frac{\pi}{4}}^{\frac{\pi}{2}}\frac{d(\sin x)}{\sin x}) Let (u = \sin x), (du=\cos xdx) (\int_{\frac{\pi}{4}}^{\frac{\pi}{2}}\frac{\cos x}{\sin x}dx=\int_{\frac{\sqrt{2}}{2}}^{1}\frac{du}{u}) (=\left[\ln u\right]{\frac{\sqrt{2}}{2}}^{1}=\ln1-\ln\frac{\sqrt{2}}{2}) (=-\left(\ln\sqrt{2}-\ln2\right)=-\frac{1}{2}\ln2+\ln2=\frac{1}{2}\ln2=\ln\sqrt{2}) (No! (\int{\frac{\pi}{4}}^{\frac{\pi}{2}}\frac{\cos x}{\sin x}dx) (\int_{\frac{\pi}{4}}^{\frac{\pi}{2}}\frac{\cos x}{\sin x}dx=\int_{\frac{\pi}{4}}^{\frac{\pi}{2}}\frac{d(\sin x)}{\sin x}) Let (u=\sin x), (du=\cos xdx) (\int_{\frac{\pi}{4}}^{\frac{\pi}{2}}\frac{\cos x}{\sin x}dx=\int_{\frac{\sqrt{2}}{2}}^{1}\frac{du}{u}) (=\left[\ln u\right]{\frac{\sqrt{2}}{2}}^{1}=\ln1-\ln\frac{\sqrt{2}}{2}) (=-\ln\frac{\sqrt{2}}{2}=\ln\frac{2}{\sqrt{2}}=\ln\sqrt{2}) (No! (\int{\frac{\pi}{4}}^{\frac{\pi}{2}}\frac{\cos x}{\sin x}dx) (\int_{\frac{\pi}{4}}^{\frac{\pi}{2}}\frac{\cos x}{\sin x}dx=\int_{\frac{\pi}{4}}^{\frac{\pi}{2}}\frac{d(\sin x)}{\sin x}) Let (u=\sin x), (du=\cos xdx) (\int_{\frac{\pi}{4}}^{\frac{\pi}{2}}\frac{\cos x}{\sin x}dx=\int_{\frac{\sqrt{2}}{2}}^{1}\frac{du}{u}) (=\left[\ln u\right]{\frac{\sqrt{2}}{2}}^{1}=\ln1-\ln\frac{\sqrt{2}}{2}) (=-\left(\frac{1}{2}\ln2-\ln2\right)=\frac{1}{2}\ln2=\ln\sqrt{2}) (No! (\int{\frac{\pi}{4}}^{\frac{\pi}{2}}\frac{\cos x}{\sin x}dx) (\int_{\frac{\pi}{4}}^{\frac{\pi}{2}}\frac{\cos x}{\sin x}dx=\int_{\frac{\pi}{4}}^{\frac{\pi}{2}}\frac{d(\sin x)}{\sin x}) Let (u=\sin x), (du=\cos xdx) (\int_{\frac{\