Math

QuestionSolve on separate sheets with signatures. No calculators. Justify answers.
1. (a) Find the integral: xarctan(x)dx\int x \arctan (x) d x (b) Evaluate: 0π/2cos(t)sin(t)dt\int_{0}^{\pi / 2} \frac{\cos (t)}{\sqrt{\sin (t)}} d t

Studdy Solution

STEP 1

Assumptions1. We know the basic rules of integration. . We understand the concept of indefinite and improper integrals.
3. We know the integration by parts formula udv=uvvdu\int u dv = uv - \int v du.
4. We understand the concept of trigonometric functions and their properties.

STEP 2

Let's start with part (a). We will use the integration by parts formula. Let's choose u=arctan(x)u = \arctan(x) and dv=xdxdv = x dx.

STEP 3

We need to calculate the derivatives and integrals for uu and dvdv.du=11+x2dxdu = \frac{1}{1 + x^2} dxv=12x2v = \frac{1}{2} x^2

STEP 4

Substitute uu, vv, dudu, and dvdv into the integration by parts formula.
xarctan(x)dx=uvvdu\int x \arctan (x) dx = u v - \int v du

STEP 5

Substitute the values of uu, vv, and dudu into the equation.
xarctan(x)dx=arctan(x)12x212x211+x2dx\int x \arctan (x) dx = \arctan(x) \cdot \frac{1}{2} x^2 - \int \frac{1}{2} x^2 \cdot \frac{1}{1 + x^2} dx

STEP 6

implify the integral.
xarctan(x)dx=12x2arctan(x)12x21+x2dx\int x \arctan (x) dx = \frac{1}{2} x^2 \arctan(x) - \frac{1}{2} \int \frac{x^2}{1 + x^2} dx

STEP 7

The integral x21+x2dx\int \frac{x^2}{1 + x^2} dx can be simplified by the substitution t=1+x2t =1 + x^2, dt=2xdxdt =2x dx.

STEP 8

Substitute tt and dtdt into the integral and simplify.
x21+x2dx=12dt=12t+C\int \frac{x^2}{1 + x^2} dx = \frac{1}{2} \int dt = \frac{1}{2} t + C

STEP 9

Substitute tt back into the equation.
x2+x2dx=2(+x2)+C\int \frac{x^2}{ + x^2} dx = \frac{}{2} ( + x^2) + C

STEP 10

Substitute this result back into the original equation.
xarctan(x)dx=2x2arctan(x)4(+x2)+C\int x \arctan (x) dx = \frac{}{2} x^2 \arctan(x) - \frac{}{4} ( + x^2) + C

STEP 11

Now, let's move to part (b). We will use the substitution t=π/ut = \pi / - u, dt=dudt = -du.

STEP 12

Substitute tt and dtdt into the integral.
0π/2cos(t)sin(t)dt=π/20cos(π/2u)sin(π/2u)du\int_{0}^{\pi /2} \frac{\cos (t)}{\sqrt{\sin (t)}} dt = - \int_{\pi /2}^{0} \frac{\cos (\pi /2 - u)}{\sqrt{\sin (\pi /2 - u)}} du

STEP 13

Use the trigonometric identities cos(π/2u)=sinu\cos (\pi /2 - u) = \sin u and sin(π/2u)=cosu\sin (\pi /2 - u) = \cos u.
π/20cos(π/2u)sin(π/2u)du=π/20sinucosudu- \int_{\pi /2}^{0} \frac{\cos (\pi /2 - u)}{\sqrt{\sin (\pi /2 - u)}} du = - \int_{\pi /2}^{0} \frac{\sin u}{\sqrt{\cos u}} du

STEP 14

Change the limits of integration.
π/20sinucosudu=0π/2sinucosudu- \int_{\pi /2}^{0} \frac{\sin u}{\sqrt{\cos u}} du = \int_{0}^{\pi /2} \frac{\sin u}{\sqrt{\cos u}} du

STEP 15

This integral is a standard integral and its value is2.
0π/2sinucosudu=2\int_{0}^{\pi /2} \frac{\sin u}{\sqrt{\cos u}} du =2So, the solutions are(a) xarctan(x)dx=2x2arctan(x)4(+x2)+C\int x \arctan (x) dx = \frac{}{2} x^2 \arctan(x) - \frac{}{4} ( + x^2) + C(b) 0π/2cos(t)sin(t)dt=2\int_{0}^{\pi /2} \frac{\cos (t)}{\sqrt{\sin (t)}} dt =2

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