Find $\displaystyle \int {x^2\,\sin x^3}{\;dx}$ by using substitution.
$\displaystyle -{{\cos x^3}\over{3}} + C$ $\displaystyle -{{\sin x^3}\over{3}} + C$ $\displaystyle {{\sin x^3}\over{3}} + C$ $\displaystyle {{\cos x^3}\over{3}} + C$
Find $\displaystyle \int {{{\sqrt{\ln x}}\over{x}}}{\;dx}$ by substituting $\displaystyle u=\ln x$
$\displaystyle \int {\sqrt{u}}{\;du} = {{2\,\left(\ln x\right)^{{{3}\over{2}}}}\over{3}} + C$ $\displaystyle {{\int {u}{\;du}}\over{2}} = {{\left(\ln x\right)^2}\over{4}} + C$ $\displaystyle \int {e^{u}}{\;du} = x + C$ $\displaystyle \int {{{1}\over{\sqrt{u}}}}{\;du} = 2\,\sqrt{\ln x} + C$
Evaluate $\displaystyle \int_{1}^{4}{{{\sqrt{x}+1}\over{\sqrt{x}}}\;dx}$ .
$\displaystyle\Big[ x+2\,\sqrt{x} \Big]_1^4 = 5$ $\displaystyle\Big[ {{2\,x^{{{3}\over{2}}}}\over{3}}+x \Big]_1^4 = {{23}\over{3}}$ $\displaystyle\Big[ 2\,\sqrt{x} \Big]_1^4 = 2$ $\displaystyle\Big[ {{\left(\sqrt{x}+1\right)^2}\over{x}} \Big]_1^4 = -{{7}\over{4}}$
Find $\displaystyle \int {x^2\,\left(x^3+3\right)^5}{\;dx}$ by substituting $\displaystyle u=x^3+3$
$\displaystyle {{\int {\left(u-3\right)\,u^5}{\;du}}\over{3}} = {{\left(x^3+3\right)^7}\over{21}}-{{\left(x^3+3\right)^6}\over{6}} + C$ $\displaystyle \int {u^4}{\;du} = {{\left(x^3+3\right)^5}\over{5}} + C$ $\displaystyle {{\int {u^5}{\;du}}\over{3}} = {{\left(x^3+3\right)^6}\over{18}} + C$ $\displaystyle \int {u^5}{\;du} = {{\left(x^3+3\right)^6}\over{6}} + C$
Find $\displaystyle \int {\sqrt{\sec x}\,\tan x}{\;dx}$ by using substitution.
$\displaystyle 2\,\sqrt{\cos x} + C$ $\displaystyle -2\,\sqrt{\cos x} + C$ $\displaystyle {{2\,\left(\cos x\right)^{{{3}\over{2}}}}\over{3}} + C$ $\displaystyle {{2}\over{\sqrt{\cos x}}} + C$
Evaluate $\displaystyle \int_{1}^{4}{x^{{{3}\over{2}}}\;dx}$ .
$\displaystyle\Big[ {{2\,x^{{{5}\over{2}}}}\over{5}} \Big]_1^4 = {{62}\over{5}}$ $\displaystyle\Big[ {{5\,x^{{{3}\over{2}}}}\over{2}} \Big]_1^4 = {{35}\over{2}}$ $\displaystyle\Big[ {{3\,\sqrt{x}}\over{2}} \Big]_1^4 = {{3}\over{2}}$ $\displaystyle\Big[ x^{{{5}\over{2}}} \Big]_1^4 = 31$ $\displaystyle\Big[ x^{{{3}\over{2}}} \Big]_1^4 = 7$
Find $\displaystyle \int {e^{x}\,\left(e^{x}+1\right)^2}{\;dx}$ by substituting $\displaystyle u=e^{x}+1$
$\displaystyle -\int {u^2}{\;du} = -{{\left(e^{x}+1\right)^3}\over{3}} + C$ $\displaystyle \int {u^2}{\;du} = {{\left(e^{x}+1\right)^3}\over{3}} + C$ $\displaystyle \int {\left(u-1\right)\,u^2}{\;du} = {{\left(e^{x}+1\right)^4}\over{4}}-{{\left(e^{x}+1\right)^3}\over{3 }} + C$ $\displaystyle \int {u}{\;du} = {{\left(e^{x}+1\right)^2}\over{2}} + C$
Find $\displaystyle \int {{{1-\sin ^2x}\over{\sin ^2x}}}{\;dx}$ .
$\displaystyle \cot x + C$ $\displaystyle -\cot x + C$ $\displaystyle x-\cot x + C$ $\displaystyle -\cot x-x + C$
Find $\displaystyle \int {\cos ^5x\,\sin x}{\;dx}$ by substituting $\displaystyle u=\cos x$
$\displaystyle -\int {u^5\,\sqrt{1-u^2}}{\;du} = {{\cos ^4x\,\left(1-\cos ^2... ...{2}}}}\over{35}}+{{8\, \left(1-\cos ^2x\right)^{{{3}\over{2}}}}\over{105}} + C$ $\displaystyle \int {u^5}{\;du} = {{\cos ^6x}\over{6}} + C$ $\displaystyle -\int {u^5}{\;du} = -{{\cos ^6x}\over{6}} + C$ $\displaystyle -\int {u^4}{\;du} = -{{\cos ^5x}\over{5}} + C$
Find $\displaystyle \int {1-{{1}\over{x^2+1}}}{\;dx}$ .
$\displaystyle x+{{1}\over{x}} + C$ $\displaystyle x-\arctan x + C$ $\displaystyle {{2\,x}\over{\left(x^2+1\right)^2}} + C$ $\displaystyle {{\ln \left(x+1\right)}\over{2}}+x-{{\ln \left(x-1\right)}\over{2 }} + C$

Department of Mathematics
Last modified: 2025-11-03