Problem (23) : Integral


\int [tan2(2x-1) + sec(2x-1)]2 dx

= \int [tan2(2x-1) tan2(2x-1) + 2 tan2(2x-1) sec(2x-1) + sec2(2x-1)] dx

= \int tan2(2x-1) tan2(2x-1) dx + 2 \int tan2(2x-1) sec(2x-1) dx + \int sec2(2x-1) dx

= \int (sec2(2x-1) – 1) tan2(2x-1)) dx + 2 \int (sec2(2x-1) – 1)sec(2x-1) dx + \int sec2(2x-1) dx

= \int sec2(2x-1) tan2(2x-1) dx – \int tan2(2x-1) dx + 2 \int sec3(2x-1) dx – 2 \int sec(2x-1) dx + \int sec2(2x-1) dx

perhatikan :

 .

\int sec2(2x-1) tan2(2x-1) dx

misal : u = tan(2x-1) \Rightarrow du = 2 sec2(2x-1) dx

= \frac{1}{2} \int u2 du

= \frac{1}{6} u3

\int sec2(2x-1) tan2(2x-1) dx = \frac{1}{6} tan3(2x-1)

 .

\int tan2(2x-1) dx = \frac{1}{2} tan (2x-1) – \frac{1}{2} (2x-1) [bukti]

 .

\int sec2(2x-1) dx = \frac{1}{2} tan(2x-1) [bukti]

 .

\int sec3(2x-1) dx = \int sec2(2x-1) sec x dx

misal : u = sec(2x-1) \Rightarrow du = 2 sec(2x-1) tan(2x-1) dx [bukti]

dv = sec2(2x-1) dx \Rightarrow v = \frac{1}{2} tan(2x-1)

= \frac{1}{2} sec(2x-1) tan(2x-1) – \int tan2(2x-1) sec(2x-1) dx

= \frac{1}{2} sec(2x-1) tan(2x-1) – \int (sec2(2x-1) – 1)sec(2x-1) dx

= \frac{1}{2} sec(2x-1) tan(2x-1) – \int sec3(2x-1) dx – \int sec(2x-1) dx

2 \int sec3(2x-1) dx = \frac{1}{2} sec(2x-1) tan(2x-1) – \frac{1}{2} ln|sec(2x-1) –tan(2x-1)|

 .

\int sec(2x-1) = \frac{1}{2} ln|sec(2x-1) – tan(2x-1)| [bukti]

 .

\int [tan2(2x-1) + sec(2x-1)]2 dx = \int sec2(2x-1) tan2(2x-1) dx – \int tan2(2x-1) dx + 2 \int sec3(2x-1) dx – 2 \int sec(2x-1) dx + \int sec2(2x-1) dx

= [\frac{1}{6} tan3(2x-1)] – [\frac{1}{2} tan (2x-1) – \frac{1}{2} (2x-1)] + [\frac{1}{2} sec(2x-1) tan(2x-1) – \frac{1}{2} ln|sec(2x-1) – tan(2x-1)|] – 2[\frac{1}{2} ln|sec(2x-1) –tan(2x-1)|] + [\frac{1}{2} tan(2x-1)]

= \frac{1}{6} tan3(2x-1) – \frac{1}{2} tan (2x-1) – \frac{1}{2} (2x-1) + \frac{1}{2} sec(2x-1) tan(2x-1) – \frac{1}{2} ln|sec(2x-1) – tan(2x-1)| – ln|sec(2x-1) –tan(2x-1)| + \frac{1}{2} tan(2x-1)

= \frac{1}{6} tan3(2x-1) + \frac{1}{2} sec(2x-1) tan(2x-1) – \frac{3}{2} ln|sec(2x-1) – tan(2x-1)| – \frac{1}{2} (2x-1)

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