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Mathematics 9 Online
OpenStudy (anonymous):

Find the indefinite integral of cos(x)(sin(sinx))

OpenStudy (anonymous):

@eliassaab

OpenStudy (anonymous):

\[I=\int\limits \cos x \sin \left( \sin x \right)dx\] put sin ~x=t cos~`x~dx=dt \[I=\int\limits \sin t~dt=-\cos ~t+c\] replace the value of t.

OpenStudy (anonymous):

@surjithayer which method did you use? integration by parts or what?

OpenStudy (anonymous):

no integration by parts ,only substitution.

OpenStudy (anonymous):

@surjithayer Thanks very much. Btw do you know how to find the indefinite integral of sqrt(1-sinx)?

OpenStudy (anonymous):

\[I=\int\limits \sqrt{1-\sin x}~dx=\int\limits \left(\sqrt{\cos ^2\frac{ x }{ 2 }+\sin ^2\frac{ x }{2 }-2\sin \frac{ x }{ 2 }\cos \frac{ x }{ 2 }}\right)dx+c\] \[=\int\limits \left( \cos \frac{ x }{ 2 }-\sin \frac{ x }{ 2} \right)dx+c\] i think now you can complete it.

OpenStudy (anonymous):

i am going to dinner.

OpenStudy (anonymous):

@surjithayer can you please explain to me how 1-sinx = to cos^2x/2 + sin^2....... the whole thing under the root?

OpenStudy (anonymous):

\[\cos ^2x+\sin ^2x=1,\sin 2x=2\sin x \cos x\]

OpenStudy (anonymous):

okay and then......

OpenStudy (anonymous):

\[\left( a-b \right)^2=a^2+b^2-2ab\]

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