4.7E: Exercises
- Page ID
- 13774
This page is a draft and is under active development.
Exercise \(\PageIndex{1}\)
In the following exercises, use a change of variables to evaluate the definite integral.
1. \(\displaystyle∫^1_0x\sqrt{1−x^2}dx\)
2. \(\displaystyle∫^1_0\frac{x}{\sqrt{1+x^2}}dx\)
- Answer
-
\(\displaystyleu=1+x^2,du=2xdx,\frac{1}{2}∫^2_1u^{−1/2}du=\sqrt{2}−1\)
3. \(\displaystyle∫^2_0\frac{t}{\sqrt{5+t^2}}dt\)
4. \(\displaystyle∫^1_0\frac{t}{\sqrt{1+t^3}}dt\)
- Answer
-
\(\displaystyleu=1+t^3,du=3t^2,\frac{1}{3}∫^2_1u^{−1/2}du=\frac{2}{3}(\sqrt{2}−1)\)
5. \(\displaystyle∫^{π/4}_0sec^2θtanθdθ\)
6. \(\displaystyle∫^{π/4}_0\frac{sinθ}{cos^4θ}dθ\)
- Answer
-
\(\displaystyleu=cosθ,du=−sinθdθ,∫^1_{1/\sqrt{2}}u^{−4}du=\frac{1}{3}(2\sqrt{2}−1)\)
Exercise \(\PageIndex{2}\)
In the following exercises, evaluate the definite integral.
1. \(\displaystyle ∫^2_1\frac{1+2x+x^2}{3x+3x^2+x^3}dx\)
- Answer
-
\(\displaystyle \frac{1}{3}ln(\frac{26}{7})\)
2. \(\displaystyle ∫^{π/4}_0tanxdx\)
3. \(\displaystyle ∫^{π/3}_0\frac{sinx−cosx}{sinx+cosx}dx\)
- Answer
-
\(\displaystyle ln(\sqrt{3}−1)\)
4. \(\displaystyle ∫^{π/2}_{π/6}cscxdx\)
5. \(\displaystyle ∫^{π/3}_{π/4}cotxdx\)
- Answer
-
\(\displaystyle \frac{1}{2}ln\frac{3}{2}\)
Exercise \(\PageIndex{3}\)
In the following exercises, use a change of variables to show that each definite integral is equal to zero.
- \(\displaystyle∫^π_0cos^2(2θ)sin(2θ)dθ\)
- \(\displaystyle∫^\sqrt{π}_0tcos(t^2)sin(t^2)dt\)
- Answer
-
\(\displaystyle u=sin(t^2);\) the integral becomes \(\displaystyle\frac{1}{2}∫^0_0udu.\)
3. \(\displaystyle∫^1_0(1−2t)dt\)
4. \(\displaystyle∫^1_0\frac{1−2t}{(1+(t−\frac{1}{2})^2)}dt\)
- Answer
-
\(\displaystyleu=(1+(t−\frac{1}{2})^2);\) the integral becomes \(\displaystyle−∫^{5/4}_{5/4}\frac{1}{u}du\).
5. \(\displaystyle∫^π_0sin((t−\frac{π}{2})^3)cos(t−\frac{π}{2})dt\)
6. \(\displaystyle∫^2_0(1−t)cos(πt)dt\)
- Answer
-
\(\displaystyle∫^{−1}_1ucos(π(1−u))du=∫^{−1}_1u[cosπcosu−sinπsinu]du=−∫^{−1}_1ucosudu=∫^{1−}_1ucosudu=0\)
since the integrand is odd.
7. \(\displaystyle∫^{3π/4}_{π/4}sin^2tcostdt\)