Question

I am practicing integration (and practicing latex). If I am wrong correct me pliz, but don't make it your practice instead of mine....

Answer 1

\[\int\limits_{ }^{ }e^x \sin(x)~dx\]integration by parts. I am choosing to differentiate the sin(x).
\[\int\limits_{ }^{ }e^x \sin(x)~dx=e^{x}\sin(x)-\int\limits_{ }^{ }e^x \cos(x)~dx\]again integration by parts, and I am differentiating cos(x) this time.\[\int\limits_{ }^{ }e^x \sin(x)~dx=e^{x}\sin(x)-e^{x}\cos(x)-\int\limits_{ }^{ }e^x \sin(x)~dx\](I get minus the last integ. because cos(x) gives a derivative of negative sin(x) then there is a minus in front of our integral in the second step, and integration by parts also leaves a negative integral. 3 MINUSES)
\[\int\limits_{ }^{ }e^x \sin(x)~dx=e^{x}\sin(x)-e^{x}\cos(x)-\int\limits_{ }^{ }e^x \sin(x)~dx\]\[2\int\limits_{ }^{ }e^x \sin(x)~dx=e^{x}\sin(x)-e^{x}\cos(x)\]\[\int\limits_{ }^{ }e^x \sin(x)~dx=\frac{1}{2}\left(\begin{matrix} e^x \sin(x)-e^x \cos(x)\end{matrix}\right) \]

Answer 2

this is the answer I got:)

Answer 3

The next one is:
\[\int\limits_{ }^{ } \sin^{-1}x~dx\]doing by parts, we get:
\[\int\limits_{ }^{ } \sin^{-1}x~dx=x \sin^{-1}x-\int\limits_{ }^{ } \frac{x}{\sqrt{1-x^2}}~dx \]\[u=1-x^2,~~~~~~~~~~~-\frac{1}{2}~du=x~dx\]\[\int\limits_{ }^{ } \sin^{-1}x~dx=x \sin^{-1}x+\frac{1}{2}\int\limits_{ }^{ } \frac{1}{\sqrt{u}}~du \]\[\int\limits_{ }^{ } \sin^{-1}x~dx=x \sin^{-1}x+\sqrt{u}+C\]\[\int\limits_{ }^{ } \sin^{-1}x~dx=x \sin^{-1}x+\sqrt{1-x^2}+C\]

Answer 4

this is my answer.