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Answer to Question #3367 in Calculus for noor

Question #3367
Evaluate the integrals :

1- sec x dx

2- x square root x+1
Expert's answer
<img src="/cgi-bin/mimetex.cgi?%5Cint%7B%5Csec%20x%20dx%7D%20=%20%5Cint%20%5Cfrac%7Bdx%7D%7B%5Csin%20%7Bx%7D%7D%20=%20%5Cint%20%5Cfrac%7B2dx%7D%7B%5Csin%20%7B%5Cfrac%7Bx%7D%7B2%7D%7D%20%5Ccos%7B%5Cfrac%7Bx%7D%7B2%7D%7D%7D%20=%20%5Cint%20%5Cfrac%7B2%5Ccos%7B%5Cfrac%7Bx%7D%7B2%7D%7Ddx%7D%7B%5Csin%20%7B%5Cfrac%7Bx%7D%7B2%7D%7D%20%5Ccos%5E2%7B%5Cfrac%7Bx%7D%7B2%7D%7D%7D%20=%20%5C%5C%20=%204%20%5Cint%20%5Cfrac%7Bd%5Csin%7B%5Cfrac%7Bx%7D%7B2%7D%7D%7D%20%7B%5Csin%20%7B%5Cfrac%7Bx%7D%7B2%7D%7D%20%5Csqrt%7B1-%5Csin%5E2%20%7B%5Cfrac%7Bx%7D%7B2%7D%7D%7D%7D%20=%20[%5Csin%20%7B%5Cfrac%7Bx%7D%7B2%7D%7D%20=%20t]%20=%20%5C%5C=%204%20%5Cint%7B%5Cfrac%20%7Bdt%7D%7Bt%5Csqrt%7B1-t%5E2%7D%7D%20%7D%20=%204%5Cint%20%5Cleft%20%28%20%5Cfrac%7B1%7D%7Bt%7D%20+%20%5Cfrac%7Bt%7D%7B1-t%5E2%7D%20%5Cright%20%29%20dt%20=%204%20%5Cln%7Bt%7D%20-%202%20%5Cln%281-t%5E2%29%20=%20%5C%5C%20=%204%20%5Cln%7B%5Csin%20%7B%5Cfrac%7Bx%7D%7B2%7D%7D%7D%20-%204%20%5Cln%7B%5Ccos%20%7B%5Cfrac%7Bx%7D%7B2%7D%7D%7D=%204%20%5Cln%20%7B%5Ctan%20%7Bx/2%7D%7D%5C%5C%20%5C%5C%20%5C%5C%20%5Cint%20%7Bx%20%5Csqrt%7B1+x%7Ddx%7D%20=%20%5Cint%7B%28%5Csqrt%7B1%20+%20x%7D+%20x%20%5Csqrt%7B1+x%7D%29dx%7D%20-%20%5Cint%5Csqrt%7B1+x%7D%20dx%20=%20%5C%5C%20=%20%5Cint%281+x%29%5E%7B3/2%7Ddx%20-%5Cint%281+x%29%5E%7B1/2%7Ddx%20=%5Cfrac%7B2%7D%7B5%7D%28x+1%29%5E%7B%5Cfrac%7B5%7D%7B2%7D%7D%20-%20%5Cfrac%7B2%7D%7B3%7D%28x+1%29%5E%7B%5Cfrac%7B3%7D%7B2%7D%7D" title="\int{\sec x dx} = \int \frac{dx}{\sin {x}} = \int \frac{2dx}{\sin {\frac{x}{2}} \cos{\frac{x}{2}}} = \int \frac{2\cos{\frac{x}{2}}dx}{\sin {\frac{x}{2}} \cos^2{\frac{x}{2}}} = \\ = 4 \int \frac{d\sin{\frac{x}{2}}} {\sin {\frac{x}{2}} \sqrt{1-\sin^2 {\frac{x}{2}}}} = [\sin {\frac{x}{2}} = t] = \\= 4 \int{\frac {dt}{t\sqrt{1-t^2}} } = 4\int \left ( \frac{1}{t} + \frac{t}{1-t^2} \right ) dt = 4 \ln{t} - 2 \ln(1-t^2) = \\ = 4 \ln{\sin {\frac{x}{2}}} - 4 \ln{\cos {\frac{x}{2}}}= 4 \ln {\tan {x/2}}\\ \\ \\ \int {x \sqrt{1+x}dx} = \int{(\sqrt{1 + x}+ x \sqrt{1+x})dx} - \int\sqrt{1+x} dx = \\ = \int(1+x)^{3/2}dx -\int(1+x)^{1/2}dx =\frac{2}{5}(x+1)^{\frac{5}{2}} - \frac{2}{3}(x+1)^{\frac{3}{2}}">

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