Homework 1 - u-Substitution and Integration By Parts

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1 :

Evaluate $\displaystyle \int \frac {1}{\cos ^2(x)\tan (x)}dx$ .

$\displaystyle \int \frac {1}{\cos ^2(x)\tan (x)}dx = \answer {\ln (|\tan (x)|)} + C$

2 :

Compute $\displaystyle \int _0^1 (2x+5)e^{x^2+5x+6}dx$ (leave your answer in terms of $e$ ).

$\displaystyle \int _0^1 (2x+5)e^{x^2+5x+6}dx = \answer {e^{12}-e^6}$

3 :

Compute $\displaystyle \int \arcsin (x)dx$ .

$\displaystyle \int \arcsin (x)dx = \answer {x\arcsin (x)+\sqrt {1-x^2}} + C$

4 :

Calculate $\displaystyle \int x^2\cos (x)dx$ .

$\displaystyle \int x^2\cos (x)dx = \answer {x^2\sin (x)+2x\cos (x)-2\sin (x)} + C$

5 :

Compute $\displaystyle \int t^2\sqrt {t-2}dt$ .

$\displaystyle \int t^2\sqrt {t-2}dt = \answer {\frac {2(t-2)^{7/2}}{7} + \frac {8(t-2)^{5/2}}{5} + \frac {8(t-2)^{3/2}}{3}} + C$

6 :

Find the general antiderivative of $e^{-2x}\sin (2x)$ with respect to $x$ .

$\answer {-\frac {e^{-2x}(\cos (2x)+\sin (2x))}{4}} + C$

7 :

Suppose a particle has velocity $v(t) = t^2e^{-t}$ . Find the total distance traveled $D$ by the particle from $t = 0$ to $t = 5$ (leave your answer in terms of $e$ ).

$D = \answer {2 - \frac {37}{e^5}}$

8 :

Calculate $\displaystyle \int \sin (\ln (x))dx$ .

$\displaystyle \int \sin (\ln (x))dx = \answer {\frac {x}{2}(\sin (\ln (x))-\cos (\ln (x)))} + C$

9 :

Compute $\displaystyle \int _{-1}^1 \arctan (x)dx$ .

$\displaystyle \int _{-1}^1 \arctan (x)dx = \answer {0}$

10 :

Find the general antiderivative of $e^{2x}\cos (\pi )$ with respect to $x$ .

$\answer {\frac {-e^{2x}}{2}} + C$

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