Evaluate $\displaystyle \int_{0}^{\pi}{\sin x-4\,\cos x\;dx}$ .
$\displaystyle\Big[ \sin x-4\,\cos x \Big]_0^\pi = 8$ $\displaystyle\Big[ -4\,\sin x-\cos x \Big]_0^\pi = 2$ $\displaystyle\Big[ \cos x-4\,\sin x \Big]_0^\pi = -2$ $\displaystyle\Big[ 4\,\sin x-\cos x \Big]_0^\pi = 2$ $\displaystyle\Big[ 4\,\sin x+\cos x \Big]_0^\pi = -2$
Find $\displaystyle \int {\cos x\,\sin ^3x}{\;dx}$ by substituting $\displaystyle u=\sin x$
$\displaystyle \int {u^3}{\;du} = {{\sin ^4x}\over{4}} + C$ $\displaystyle \int {u^2}{\;du} = {{\sin ^3x}\over{3}} + C$ $\displaystyle \int {u^3\,\sqrt{1-u^2}}{\;du} = -{{\sin ^2x\,\left(1-\sin ^2... ...er{2}}}}\over{5}}-{{2 \,\left(1-\sin ^2x\right)^{{{3}\over{2}}}}\over{15}} + C$ $\displaystyle -\int {u^3}{\;du} = -{{\sin ^4x}\over{4}} + C$
Find $\displaystyle \int {x\,\cos x^2}{\;dx}$ by using substitution.
$\displaystyle {{\cos x^2}\over{2}} + C$ $\displaystyle -{{\cos x^2}\over{2}} + C$ $\displaystyle -{{\sin x^2}\over{2}} + C$ $\displaystyle {{\sin x^2}\over{2}} + C$
Find $\displaystyle \int {{{1}\over{3\,x+2}}}{\;dx}$ by using substitution.
$\displaystyle -{{\ln \left(3\,x+2\right)}\over{3}} + C$ $\displaystyle {{\ln \left(3\,x+2\right)}\over{3}} + C$ $\displaystyle -\ln \left(3\,x+2\right) + C$ $\displaystyle \ln \left(3\,x+2\right) + C$ $\displaystyle -{{1}\over{3\,x+2}} + C$
Evaluate $\displaystyle \int_{1}^{4}{{{1}\over{\sqrt{x}}}\;dx}$ .
$\displaystyle\Big[ {{1}\over{2\,\sqrt{x}}} \Big]_1^4 = -{{1}\over{4}}$ $\displaystyle\Big[ {{1}\over{\sqrt{x}}} \Big]_1^4 = -{{1}\over{2}}$ $\displaystyle\Big[ 2\,\sqrt{x} \Big]_1^4 = 2$ $\displaystyle\Big[ -{{1}\over{2\,x^{{{3}\over{2}}}}} \Big]_1^4 = {{7}\over{16}}$ $\displaystyle\Big[ \sqrt{x} \Big]_1^4 = 1$
Find $\displaystyle \int {{{x}\over{\sqrt{x+1}}}}{\;dx}$ by substituting u = x + 1
$\displaystyle \int {{{u+1}\over{\sqrt{u}}}}{\;du} = {{2\,\left(x-1\right)^{{{3}\over{2}}}}\over{3}}+2\,\sqrt{x-1} + C$ $\displaystyle \int {{{1}\over{\sqrt{u}}}}{\;du} = 2\,\sqrt{x-1} + C$ $\displaystyle \int {{{1}\over{\sqrt{u}}}}{\;du} = 2\,\sqrt{x+1} + C$ $\displaystyle \int {{{u-1}\over{\sqrt{u}}}}{\;du} = {{2\,\left(x+1\right)^{{{3}\over{2}}}}\over{3}}-2\,\sqrt{x+1} + C$
Find $\displaystyle \int {{{1}\over{x\,\sqrt{\ln x}}}}{\;dx}$ by substituting $\displaystyle u=\ln x$
$\displaystyle \int {e^{u}}{\;du} = x + C$ $\displaystyle \int {{{1}\over{u^{{{3}\over{2}}}}}}{\;du} = -{{2}\over{\sqrt{\ln x}}} + C$ $\displaystyle \int {{{1}\over{\sqrt{u}}}}{\;du} = 2\,\sqrt{\ln x} + C$ $\displaystyle -{{\int {u}{\;du}}\over{2}} = -{{\left(\ln x\right)^2}\over{4}} + C$
Find $\displaystyle -\int {e^ {- x }\,\left(e^ {- x }+1\right)^2}{\;dx}$ by substituting $\displaystyle u=e^ {- x }+1$
$\displaystyle \int {u}{\;du} = {{\left(e^ {- x }+1\right)^2}\over{2}} + C$ $\displaystyle \int {u^2}{\;du} = {{\left(e^ {- x }+1\right)^3}\over{3}} + C$ $\displaystyle \int {\left(u-1\right)\,u^2}{\;du} = {{\left(e^ {- x }+1\right)^4}\over{4}}-{{\left(e^ {- x }+1\right)^3 }\over{3}} + C$ $\displaystyle -\int {u^2}{\;du} = -{{\left(e^ {- x }+1\right)^3}\over{3}} + C$
Evaluate $\displaystyle \int_{0}^{1}{\left(2-x^2\right)^2\;dx}$ .
$\displaystyle\Big[ {{x^3}\over{3}} \Big]_0^1 = {{1}\over{3}}$ $\displaystyle\Big[ 2\,x-{{x^3}\over{3}} \Big]_0^1 = {{5}\over{3}}$ $\displaystyle\Big[ {{x^5}\over{5}}-{{4\,x^3}\over{3}}+4\,x \Big]_0^1 = {{43}\over{15}}$ $\displaystyle\Big[ x^4-4\,x^2+4 \Big]_0^1 = -3$
Evaluate $\displaystyle \int_{1}^{2}{e^{x-1}\;dx}$ .
$\displaystyle\Big[ e^{x-1} \Big]_{ 1}^{ 2} = e-1$ $\displaystyle\Big[ e^{x} \Big]_{ 1}^{ 2} = e^2-e$ $\displaystyle\Big[ e^{x} \Big]_{ 1}^{ 2} = e^2-e$ $\displaystyle\Big[ -e^{x} \Big]_{ 1}^{ 2} = e-e^2$

Department of Mathematics
Last modified: 2026-05-19