Generating...                               ibee2025_n4

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

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

  2. Find $\displaystyle \int {x^2\,\sin x^3}{\;dx}$ by using substitution.

    $\displaystyle {{\sin x^3}\over{3}} + C$ $\displaystyle {{\cos x^3}\over{3}} + C$ $\displaystyle -{{\sin x^3}\over{3}} + C$ $\displaystyle -{{\cos x^3}\over{3}} + C$

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

    $\displaystyle \int {u^4}{\;du} = {{\left(x^3+3\right)^5}\over{5}} + C$ $\displaystyle {{\int {\left(u-3\right)\,u^5}{\;du}}\over{3}} = {{\left(x^3+3\right)^7}\over{21}}-{{\left(x^3+3\right)^6}\over{6}} + C$ $\displaystyle {{\int {u^5}{\;du}}\over{3}} = {{\left(x^3+3\right)^6}\over{18}} + C$ $\displaystyle \int {u^5}{\;du} = {{\left(x^3+3\right)^6}\over{6}} + C$

  4. Find $\displaystyle \int {\sin ^2x}{\;dx}$ .

    $\displaystyle {{\sin \left(4\,x\right)}\over{32}}-{{\sin \left(2\,x\right)}\over{ 4}}+{{3\,x}\over{8}} + C$ $\displaystyle \sin x-{{\sin ^3x}\over{3}} + C$ $\displaystyle {{x}\over{2}}-{{\sin \left(2\,x\right)}\over{4}} + C$ $\displaystyle {{\cos ^3x}\over{3}}-\cos x + C$ $\displaystyle {{\sin \left(2\,x\right)}\over{4}}+{{x}\over{2}} + C$ $\displaystyle {{\sin \left(4\,x\right)}\over{32}}+{{\sin \left(2\,x\right)}\over{ 4}}+{{3\,x}\over{8}} + C$

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

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


Department of Mathematics
Last modified: 2025-09-14