Generating...                               s20quiz18_n22

  1. Find $\displaystyle \int {\sqrt{x-1}\,x}{\;dx}$ by substituting u = x − 1

    $\displaystyle \int {\left(u-1\right)\,\sqrt{u}}{\;du} = {{2\,\left(x+1\right)^{{{5}\over{2}}}}\over{5}}-{{2\,\left(x+1 \right)^{{{3}\over{2}}}}\over{3}} + C$ $\displaystyle \int {\sqrt{u}\,\left(u+1\right)}{\;du} = {{2\,\left(x-1\right)^{{{5}\over{2}}}}\over{5}}+{{2\,\left(x-1 \right)^{{{3}\over{2}}}}\over{3}} + C$ $\displaystyle \int {\sqrt{u}}{\;du} = {{2\,\left(x-1\right)^{{{3}\over{2}}}}\over{3}} + C$ $\displaystyle \int {\sqrt{u}}{\;du} = {{2\,\left(x+1\right)^{{{3}\over{2}}}}\over{3}} + C$

  2. Find $\displaystyle \int {{{1}\over{x^2+1}}+1}{\;dx}$.

    $\displaystyle \arctan x+x + C$ $\displaystyle -{{\ln \left(x+1\right)}\over{2}}+x+{{\ln \left(x-1\right)}\over{ 2}} + C$ $\displaystyle x-{{1}\over{x}} + C$ $\displaystyle -{{2\,x}\over{\left(x^2+1\right)^2}} + C$

  3. Evaluate $\displaystyle \int_{1}^{4}{{{1}\over{x^2}}\;dx}$.

    $\displaystyle\Big[ {{1}\over{x}} \Big]_1^4 = -{{3}\over{4}}$ $\displaystyle\Big[ -{{2}\over{x^3}} \Big]_1^4 = {{63}\over{32}}$ $\displaystyle\Big[ -{{1}\over{x}} \Big]_1^4 = {{3}\over{4}}$ $\displaystyle\Big[ -{{1}\over{x^2}} \Big]_1^4 = {{15}\over{16}}$ $\displaystyle\Big[ {{1}\over{x^2}} \Big]_1^4 = -{{15}\over{16}}$

  4. Find $\displaystyle \int {{{\cos ^2x+1}\over{\cos ^2x}}}{\;dx}$.

    $\displaystyle \tan x+x + C$ $\displaystyle \tan x + C$ $\displaystyle \tan x-x + C$ $\displaystyle -\tan x + C$

  5. Find $\displaystyle \int {x^3\,\left(x^4+2\right)^3}{\;dx}$ by substituting $\displaystyle u=x^4+2$

    $\displaystyle {{\int {u^3}{\;du}}\over{4}} = {{\left(x^4+2\right)^4}\over{16}} + C$ $\displaystyle {{\int {\left(u-2\right)\,u^3}{\;du}}\over{4}} = {{\left(x^4+2\right)^5}\over{20}}-{{\left(x^4+2\right)^4}\over{8}} + C$ $\displaystyle \int {u^3}{\;du} = {{\left(x^4+2\right)^4}\over{4}} + C$ $\displaystyle \int {u^2}{\;du} = {{\left(x^4+2\right)^3}\over{3}} + C$

  6. Find $\displaystyle \int {\left(\sec x\right)^2\,\tan x}{\;dx}$ by using substitution.

    $\displaystyle -{{1}\over{\cos x}} + C$ $\displaystyle {{1}\over{\cos x}} + C$ $\displaystyle \ln \cos x + C$ $\displaystyle {{\tan ^2x}\over{2}} + C$

  7. Evaluate $\displaystyle \int_{0}^{{{\pi}\over{4}}}{{{1-\cos ^2x}\over{\cos ^2x}}\;dx}$.

    $\displaystyle\Big[ -\tan x \Big]_{ 0}^{ {{\pi}\over{4}}} = -1$ $\displaystyle\Big[ \tan x-x \Big]_{ 0}^{ {{\pi}\over{4}}} = -{{\pi-4}\over{4}}$ $\displaystyle\Big[ {{1}\over{\cos ^2x}}-1 \Big]_{ 0}^{ {{\pi}\over{4}}} = 1$ $\displaystyle\Big[ x-\sin x \Big]_{ 0}^{ {{\pi}\over{4}}} = {{\sqrt{2}\,\pi-4}\over{2^{{{5}\over{2}}}}}$

  8. Find $\displaystyle \int {e^{x}\,\sqrt{e^{x}+1}}{\;dx}$ by substituting $\displaystyle u=e^{x}+1$

    $\displaystyle \int {{{1}\over{\sqrt{u}}}}{\;du} = 2\,\sqrt{e^{x}+1} + C$ $\displaystyle \int {\sqrt{u}}{\;du} = {{2\,\left(e^{x}+1\right)^{{{3}\over{2}}}}\over{3}} + C$ $\displaystyle -\int {\sqrt{u}}{\;du} = -{{2\,\left(e^{x}+1\right)^{{{3}\over{2}}}}\over{3}} + C$ $\displaystyle \int {\left(u-1\right)\,\sqrt{u}}{\;du} = {{2\,\left(e^{x}+1\... ...}\over{2}}}}\over{5}}-{{2\,\left(e^{x }+1\right)^{{{3}\over{2}}}}\over{3}} + C$

  9. Find $\displaystyle \int {\cos x\,\sin ^3x}{\;dx}$ by substituting $\displaystyle u=\sin x$

    $\displaystyle \int {u^3\,\sqrt{1-u^2}}{\;du} = -{{\sin ^2x\,\left(1-\sin ^2... ...er{2}}}}\over{5}}-{{2 \,\left(1-\sin ^2x\right)^{{{3}\over{2}}}}\over{15}} + C$ $\displaystyle \int {u^2}{\;du} = {{\sin ^3x}\over{3}} + C$ $\displaystyle \int {u^3}{\;du} = {{\sin ^4x}\over{4}} + C$ $\displaystyle -\int {u^3}{\;du} = -{{\sin ^4x}\over{4}} + C$

  10. Evaluate $\displaystyle \int_{0}^{\pi}{3\,\cos x-3\,\sin x\;dx}$.

    $\displaystyle\Big[ -3\,\sin x-3\,\cos x \Big]_0^\pi = 6$ $\displaystyle\Big[ 3\,\cos x-3\,\sin x \Big]_0^\pi = -6$ $\displaystyle\Big[ 3\,\sin x+3\,\cos x \Big]_0^\pi = -6$ $\displaystyle\Big[ 3\,\sin x-3\,\cos x \Big]_0^\pi = 6$ $\displaystyle\Big[ 3\,\cos x-3\,\sin x \Big]_0^\pi = -6$


Department of Mathematics
Last modified: 2025-06-19