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

    $\displaystyle\Big[ x-{{x^3}\over{3}} \Big]_0^1 = {{2}\over{3}}$ $\displaystyle\Big[ x^4-2\,x^2+1 \Big]_0^1 = -1$ $\displaystyle\Big[ {{x^3}\over{3}} \Big]_0^1 = {{1}\over{3}}$ $\displaystyle\Big[ {{x^5}\over{5}}-{{2\,x^3}\over{3}}+x \Big]_0^1 = {{8}\over{15}}$

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

    $\displaystyle\Big[ -{{1}\over{2\,x^{{{3}\over{2}}}}} \Big]_1^4 = {{7}\over{16}}$ $\displaystyle\Big[ {{1}\over{\sqrt{x}}} \Big]_1^4 = -{{1}\over{2}}$ $\displaystyle\Big[ \sqrt{x} \Big]_1^4 = 1$ $\displaystyle\Big[ 2\,\sqrt{x} \Big]_1^4 = 2$ $\displaystyle\Big[ {{1}\over{2\,\sqrt{x}}} \Big]_1^4 = -{{1}\over{4}}$

  3. Find $\displaystyle \int {{{1}\over{x\,\sqrt{\ln x}}}}{\;dx}$ by substituting $\displaystyle u=\ln x$

    $\displaystyle -{{\int {u}{\;du}}\over{2}} = -{{\left(\ln x\right)^2}\over{4}} + C$ $\displaystyle \int {{{1}\over{u^{{{3}\over{2}}}}}}{\;du} = -{{2}\over{\sqrt{\ln x}}} + C$ $\displaystyle \int {e^{u}}{\;du} = x + C$ $\displaystyle \int {{{1}\over{\sqrt{u}}}}{\;du} = 2\,\sqrt{\ln x} + C$

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

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

  5. Find $\displaystyle \int {\sqrt{x}\,\left(x+1\right)^2}{\;dx}$.

    $\displaystyle {{2\,x^{{{9}\over{2}}}}\over{9}} + C$ $\displaystyle {{2\,x^{{{7}\over{2}}}}\over{7}}+{{4\,x^{{{5}\over{2}}}}\over{5}}+ {{2\,x^{{{3}\over{2}}}}\over{3}} + C$ $\displaystyle {{\left(x+1\right)^2}\over{2\,\sqrt{x}}}+2\,\sqrt{x}\,\left(x+1 \right) + C$ $\displaystyle {{2\,x^{{{5}\over{2}}}}\over{5}} + C$

  6. Find $\displaystyle \int {{{\cos \sqrt{x}}\over{\sqrt{x}}}}{\;dx}$ by using substitution.

    $\displaystyle 2\,\sin \sqrt{x} + C$ $\displaystyle -2\,\sin \sqrt{x} + C$ $\displaystyle -2\,\cos \sqrt{x} + C$ $\displaystyle 2\,\cos \sqrt{x} + C$

  7. Find $\displaystyle \int {\cos x\,\sin ^5x}{\;dx}$ by substituting $\displaystyle u=\sin x$

    $\displaystyle \int {u^5\,\sqrt{1-u^2}}{\;du} = -{{\sin ^4x\,\left(1-\sin ^2... ...{2}}}}\over{35}}-{{8\, \left(1-\sin ^2x\right)^{{{3}\over{2}}}}\over{105}} + C$ $\displaystyle \int {u^5}{\;du} = {{\sin ^6x}\over{6}} + C$ $\displaystyle \int {u^4}{\;du} = {{\sin ^5x}\over{5}} + C$ $\displaystyle -\int {u^5}{\;du} = -{{\sin ^6x}\over{6}} + C$

  8. Evaluate $\displaystyle \int_{-2}^{-1}{e^{x+2}\;dx}$.

    $\displaystyle\Big[ e^{x+2} \Big]_{ -2}^{ -1} = e-1$ $\displaystyle\Big[ e^{x} \Big]_{ -2}^{ -1} = e^ {- 1 }-e^ {- 2 }$ $\displaystyle\Big[ e^{x+3} \Big]_{ -2}^{ -1} = e^2-e$ $\displaystyle\Big[ 2\,e^{x} \Big]_{ -2}^{ -1} = 2\,e^ {- 1 }-2\,e^ {- 2 }$

  9. Find $\displaystyle \int {\left(\sec x\right)^3\,\tan x}{\;dx}$ by using substitution.

    $\displaystyle -{{1}\over{\cos x}} + C$ $\displaystyle {{1}\over{2\,\cos ^2x}} + C$ $\displaystyle -{{1}\over{2\,\cos ^2x}} + C$ $\displaystyle {{1}\over{3\,\cos ^3x}} + C$

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

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


Department of Mathematics
Last modified: 2025-05-07