a)&nbsp;u=x-3. Then&nbsp;du=dx.∫-211uduThe integral of&nbsp;1u&nbsp;with respect to&nbsp;u&nbsp;is&nbsp;ln(|u|).ln(|u|)]-21Evaluate&nbsp;ln(|u|)&nbsp;at&nbsp;1&nbsp;and at&nbsp;-2.ln(|1|)-ln(|-2|)Use the quotient property of logarithmslogb(x)-logb(y)=logb(xy).ln(|1||-2|)Simplify.ln(12)&nbsp;b) Let&nbsp;u=πx. Then&nbsp;du=πdx, so&nbsp;1πdu=dx∫0πtan(u)1πduCombine&nbsp;tan(u)&nbsp;and&nbsp;1π.∫0πtan(u)πduSince&nbsp;1π&nbsp;is constant with respect to&nbsp;u, move&nbsp;1π&nbsp;out of the integral.1π∫0πtan(u)duThe integral of&nbsp;tan(u)&nbsp;with respect to&nbsp;u&nbsp;is&nbsp;ln(|sec(u)|).1πln(|sec(u)|)]0πEvaluate&nbsp;ln(|sec(u)|)&nbsp;at&nbsp;π&nbsp;and at&nbsp;0.1π(ln(|sec(π)|)-ln(|sec(0)|))Simplify.ln(|sec(π)||1|)π=0

Solve my Integral Calculus problem. Here is a photo.

aiman altaf

Answered question

2022-06-05

Answer & Explanation

star233

Skilled2022-11-03Added 403 answers

a) $u = x - 3$ . Then $d u = d x$ .

$\int_{- 2}^{1} \frac{1}{u} d u$

The integral of $\frac{1}{u}$ with respect to $u$ is $\ln (| u |)$ .

$\ln (| u |)]_{- 2}^{1}$

Evaluate $\ln (| u |)$ at $1$ and at $- 2$ .

$\ln (| 1 |) - \ln (| - 2 |)$

Use the quotient property of logarithms

$\log_{b} (x) - \log_{b} (y) = \log_{b} (\frac{x}{y})$ .

$\ln (\frac{| 1 |}{| - 2 |})$

Simplify.

$\ln (\frac{1}{2})$

b) Let $u = π x$ . Then $d u = π d x$ , so $\frac{1}{π} d u = d x$

$\int_{0}^{π} \tan (u) \frac{1}{π} d u$

Combine $\tan (u)$ and $\frac{1}{π}$ .

$\int_{0}^{π} \frac{\tan (u)}{π} d u$

Since $\frac{1}{π}$ is constant with respect to $u$ , move $\frac{1}{π}$ out of the integral.

$\frac{1}{π} \int_{0}^{π} \tan (u) d u$

The integral of $\tan (u)$ with respect to $u$ is $\ln (| \sec (u) |)$ .

$\frac{1}{π} \ln (| \sec (u) |)]_{0}^{π}$

Evaluate $\ln (| \sec (u) |)$ at $π$ and at $0$ .

$\frac{1}{π} (\ln (| \sec (π) |) - \ln (| \sec (0) |))$

Simplify.

$\frac{\ln (\frac{| \sec (π) |}{| 1 |})}{π}$ =0

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