Question

how to integrate this ?

Answer 1

\[e ^{x ^{2}}\sin2xdx\]

Answer 2

and also this \[\sqrt{(1-\frac{ 1 }{ 3 }\sin ^{2}x)dx}\]

Answer 3

The first one looks like Integration by Parts (To me atleast)

Answer 4

@FortyTheRapper is correct. in 1st one we have to integrate by parts \[I = \int\limits_{}^{} e^{x^{2}} \sin2x = \int\limits e^{x^{2}} (\frac{ d (\frac{ -\cos2x }{ 2 } ) }{ dx })dx\]
therefore \[I = e^{x^{2}}(\frac{ -\cos2x }{ 2 }) - \int\limits (\frac{ -\cos2x }{ 2} ) ( \frac{ de^{x^{2}} }{ dx })dx\]\[I =- e^{x^{2}}\frac{ \cos2x }{ 2} +\frac{ 1 }{ 2 } \int\limits \cos2x ( e^{x^{2}}) (2x) dx \]andd we get\[I = -e^{x^{2}}\frac{ \cos 2x }{ 2 } + \int\limits \cos2x (x e^{x^{2}}) dx\]

Answer 5

now lets take \[I_{0} = \int\limits \cos2x (xe^{x^{2}}) dx \]

Answer 6

and to solve it lets use the substitution \[v = e^{x^{2}} \]

Answer 7

by differentiation the above we get \[\frac{ dv }{ dx} = 2x e^{x^{2}}\]\[dx = \frac{ dv }{ 2x e^{x^{2}} } = \frac{ dv }{ 2x v }\]

Answer 8

now from the substitution we can get \[I_{0} = \int\limits vx \cos 2(\sqrt{\ln v }) \times \frac{ dv }{ 2vx } = \frac{ 1 }{ 2 } \int\limits \cos 2(\sqrt{\ln v}) \ dv \]

Answer 9

here\[\cos 2x \ \ became \ \cos 2\sqrt{\ln v} \] because our substitution says \[v = e^{x^{2}}\]\[\ln v = \ln e^{x^{2}} \]\[\ln v = x^{2} \ln e\]since \[\ln e =1 \]\[x^{2} = \ln v\]\[x = \sqrt{\ln v}\]

Answer 10

now \[I_{0} = \frac{ 1 }{ 2 } \sin2(\sqrt{lnv}) \times v \times \sqrt{\ln v}\]
by substituting back we get \[I_{0} = \frac{ 1 }{ 2 } \sin2x (xe^{x^{2}})\]
which means \[I = -e^{x^{2}}\frac{ \cos2x }{ 2 } +\frac{ 1 }{ 2 } \sin2x (x e^{x^{2}}) \]

Answer 11

hope this would help u :)

Answer 12

cool :))) Thank you so much

Answer 13

Can you simplify the second equation

Answer 14

im trying bro..but still no luck on the 2nd one :(

Answer 15

Cool I just hope you could answer that