Generating...                               s20quiz18_n11

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

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

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

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

  3. Evaluate $\displaystyle \int_{0}^{\pi}{4\,\sin x-2\,\cos x\;dx}$.

    $\displaystyle\Big[ 4\,\cos x-2\,\sin x \Big]_0^\pi = -8$ $\displaystyle\Big[ 2\,\sin x-4\,\cos x \Big]_0^\pi = 8$ $\displaystyle\Big[ 2\,\sin x+4\,\cos x \Big]_0^\pi = -8$ $\displaystyle\Big[ 4\,\sin x-2\,\cos x \Big]_0^\pi = 4$ $\displaystyle\Big[ -2\,\sin x-4\,\cos x \Big]_0^\pi = 8$

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

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

  5. Find $\displaystyle \int {{{\left(\ln x\right)^2}\over{x}}}{\;dx}$ by substituting $\displaystyle u=\ln x$

    $\displaystyle \int {u}{\;du} = {{\left(\ln x\right)^2}\over{2}} + C$ $\displaystyle \int {e^{u}}{\;du} = x + C$ $\displaystyle 2\,\int {u}{\;du} = \left(\ln x\right)^2 + C$ $\displaystyle \int {u^2}{\;du} = {{\left(\ln x\right)^3}\over{3}} + C$

  6. Find $\displaystyle \int {{{x^2}\over{x-1}}}{\;dx}$ by substituting u = x − 1

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

  7. Evaluate $\displaystyle \int_{0}^{1}{\left(x+3\right)^2\;dx}$.

    $\displaystyle\Big[ x^2+6\,x+9 \Big]_0^1 = 7$ $\displaystyle\Big[ {{x^2}\over{2}}+3\,x \Big]_0^1 = {{7}\over{2}}$ $\displaystyle\Big[ {{x^2}\over{2}} \Big]_0^1 = {{1}\over{2}}$ $\displaystyle\Big[ {{x^3}\over{3}}+3\,x^2+9\,x \Big]_0^1 = {{37}\over{3}}$

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

    $\displaystyle\Big[ x^{{{5}\over{2}}} \Big]_1^4 = 31$ $\displaystyle\Big[ {{5\,x^{{{3}\over{2}}}}\over{2}} \Big]_1^4 = {{35}\over{2}}$ $\displaystyle\Big[ {{3\,\sqrt{x}}\over{2}} \Big]_1^4 = {{3}\over{2}}$ $\displaystyle\Big[ {{2\,x^{{{5}\over{2}}}}\over{5}} \Big]_1^4 = {{62}\over{5}}$ $\displaystyle\Big[ x^{{{3}\over{2}}} \Big]_1^4 = 7$

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

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

  10. Find $\displaystyle \int {\sqrt{2\,x+3}}{\;dx}$ by using substitution.

    $\displaystyle -{{2\,\left(2\,x+3\right)^{{{3}\over{2}}}}\over{3}} + C$ $\displaystyle {{2\,\left(2\,x+3\right)^{{{3}\over{2}}}}\over{3}} + C$ $\displaystyle -{{\left(2\,x+3\right)^{{{3}\over{2}}}}\over{3}} + C$ $\displaystyle {{\left(2\,x+3\right)^{{{3}\over{2}}}}\over{3}} + C$ $\displaystyle 2\,\sqrt{2\,x+3} + C$


Department of Mathematics
Last modified: 2026-03-24