Question

webwork 5 - topics 10 - 12: probl
(1 point)
evaluate the definite integral ( i=int_{0}^{pi / 2} \frac{sin (x) cos (x)}{cos ^{2}(x)+7} d x )
( i= )

Explanation:

Step1: Substitute \( u = \cos(x) \)

Differentiate \( u=\cos(x) \), we get \( du=-\sin(x)dx \). When \( x = 0 \), \( u=\cos(0) = 1 \); when \( x=\frac{\pi}{2} \), \( u=\cos(\frac{\pi}{2})=0 \). The integral \( I=\int_{0}^{\pi/2}\frac{\sin(x)\cos(x)}{\cos^{2}(x)+7}dx \) becomes \( I=-\int_{1}^{0}\frac{u}{u^{2}+7}du=\int_{0}^{1}\frac{u}{u^{2}+7}du \).

Step2: Integrate \( \frac{u}{u^{2}+7} \)

Let \( t = u^{2}+7 \), then \( dt = 2udu \), and \( udu=\frac{1}{2}dt \). When \( u = 0 \), \( t = 7 \); when \( u = 1 \), \( t=8 \). So \( I=\frac{1}{2}\int_{7}^{8}\frac{dt}{t} \).

Step3: Evaluate the integral of \( \frac{1}{t} \)

The integral of \( \frac{1}{t} \) is \( \ln|t| \). So \( I=\frac{1}{2}[\ln(t)]_{7}^{8}=\frac{1}{2}(\ln(8)-\ln(7))=\frac{1}{2}\ln(\frac{8}{7}) \).

Answer:

\(\frac{1}{2}\ln(\frac{8}{7})\)